To draw a Lewis structure that obeys the octet rule for each of the molecules and ions listed, simply follow the steps outlined above and make sure that each atom has a formal charge of zero or close to zero.
For each of the molecules and ions listed, you can draw a Lewis structure that obeys the octet rule by following these steps:
1. Identify the central atom: The first atom listed in each molecule or ion is the central atom.
2. Count the number of valence electrons: Each atom has a certain number of valence electrons based on its position in the periodic table.
3. Form single bonds: Make single bonds between the central atom and each of the other atoms to use up the available valence electrons.
4. Add lone pairs: If the central atom still has electrons remaining, add lone pairs to satisfy the octet rule.
5. Check the formal charges: Make sure all atoms have formal charges of zero or close to zero.
Let's look at each molecule and ion individually:
a. POCl3, SO42−, XeO4, PO43−, ClO4−:
POCl3: The central atom is phosphorus (P) and it has 5 valence electrons. We form single bonds between the P atom and each of the other atoms, giving P a total of 8 electrons. The formal charge of each atom is zero.
SO42−: The central atom is sulfur (S) and it has 6 valence electrons. We form single bonds between the S atom and each of the other atoms, giving S a total of 8 electrons. The formal charge of each atom is zero.
XeO4: The central atom is xenon (Xe) and it has 8 valence electrons. We form single bonds between the Xe atom and each of the other atoms, giving Xe a total of 8 electrons. The formal charge of each atom is zero.
PO43−: The central atom is phosphorus (P) and it has 5 valence electrons. We form single bonds between the P atom and each of the other atoms, giving P a total of 8 electrons. The formal charge of each atom is zero.
ClO4−: The central atom is chlorine (Cl) and it has 7 valence electrons. We form single bonds between the Cl atom and each of the other atoms, giving Cl a total of 8 electrons. The formal charge of each atom is zero.
b. NF3, SO32−, PO33−, ClO3−:
NF3: The central atom is nitrogen (N) and it has 5 valence electrons. We form single bonds between the N atom and each of the other atoms, giving N a total of 8 electrons. The formal charge of each atom is zero.
SO32−: The central atom is sulfur (S) and it has 6 valence electrons. We form single bonds between the S atom and each of the other atoms, giving S a total of 8 electrons. The formal charge of each atom is zero.
PO33−: The central atom is phosphorus (P) and it has 5 valence electrons. We form single bonds between the P atom and each of the other atoms, giving P a total of 8 electrons. The formal charge of each atom is zero.
ClO3−: The central atom is chlorine (Cl) and it has 7 valence electrons. We form single bonds between the Cl atom and each of the other atoms, giving Cl a total of 8 electrons. The formal charge of each atom is zero.
c. ClO2−, SCl2, PCl2−:
ClO2−: The central atom is chlorine (Cl) and it has 7 valence electrons. We form single bonds between the Cl atom and each of the other atoms, giving Cl a total of 8 electrons. The formal charge of each atom is zero.
SCl2: The central atom is sulfur (S) and it has 6 valence electrons. We form single bonds between the S atom and each of the other atoms, giving S a total of 8 electrons. The formal charge of each atom is zero.
PCl2−: The central atom is phosphorus (P) and it has 5 valence electrons. We form single bonds between the P atom and each of the other atoms, giving P a total of 8 electrons. The formal charge of each atom is zero.
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Given that 4 NH3 + 5 O2 → 4 NO + 6 H2O, if 3.00 mol NH3 were made to react with excess of oxygen gas, the amount of H2O formed would be
Predict the principal organic product of the following reaction. Specify stereochemistry where appropriate.
The major organic product of an SN2 substitution reaction is an alkene, which may be either in retention or inversion of configuration relative to the original substrate.
The reaction you are asking about is an SN2 substitution reaction, in which a nucleophile (Nu) displaces a leaving group (LG) from a molecule with an alkyl halide substrate. The major organic product of this reaction will be an alkene, which has the same carbon chain as the alkyl halide substrate. Depending on the relative configuration of the substrate, the alkene product may be the same as the original substrate (retention) or have its configuration inverted (inversion). If stereochemistry is relevant to the question, then it should be specified in the answer.
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Please help me. Thank you
The standard change in Gibbs energy at 25 degree Celsius is 490.6 °C. for given equilibrium partial pressure .
What is Gibbs energy ?The Gibbs energy is the thermodynamic potential that is minimized when a system reaches chemical equilibrium at constant pressure and temperature when not driven by an applied electrolytic voltage. Its derivative with respect to the reaction coordinate of the system then vanishes at the equilibrium point.
Using the formula
ΔG° = - R × T ln K
WHERE R= 8.3144598 J⋅mol⁻¹⋅K⁻¹.
T = 298 K
K = 0.82
SOLVING ,
The standard change in Gibbs energy at 25 degree Celsius is 490.6 °C.
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Which equation is a correctly written thermochemical equation?
OC3H8 (g) +502 (g) → 3CO2 (g) + 4H₂O (1), AH= -2,220 kJ/mol
OFe (s) + O2 (g) → Fe₂O3 (s), AH= -3,926 kJ
ONH₂Cl → NH₂ + + Cl
O2C8H18 + 250216CO2 + 18H₂O, AH=-5,471 kJ/mol
Answer:
The correctly written thermochemical equation is:
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l), ΔH = -2,220 kJ/mol
This equation represents the combustion of propane (C3H8) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O), with a heat release of -2,220 kJ/mol. The state symbols (g) for gases and (l) for liquids indicate the physical state of each substance at standard conditions.
Explanation:
ABOVE
Write the electronic configuration and draw the orbital diagram for the element: lead (Z=82) State if it is diamagnetic/paramagnetic. Please decide the diamagnetic/paramagnetic property based on the orbital diagram only! (It is okay to use the noble gas in square brackets here)
Answer:
See below.
Explanation:
The atomic number of lead (Pb) is 82, which means it has 82 electrons. The electronic configuration of lead is
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d¹⁰ 6s² 6p²
The orbital diagram for the valence electrons of lead (Pb) is
↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓
s s p p p p d d
2 1 6 2 6 2 10 10
|||||||||
1 2 3 4 5 6 7 8
The notation ↑↓ represents a pair of electrons with opposite spins.
To determine if lead (Pb) is diamagnetic or paramagnetic, we need to look at whether there are any unpaired electrons. Based on the orbital diagram, we can see that all the electrons in the valence shell are paired, meaning that lead (Pb) is diamagnetic.
There are 7.68 × 1025 atoms of phosphorous in how many moles of diphosphorous pentoxide?
Answer:
7.68 x 1025 atoms of phosphorous correspond to 1.06 mole of diphosphorous pentoxide. This can also be written as 1.06 mol of P2O5.
For Mn3+, write an equation that shows how the cation acts as an acid. express your answer as a chemical equation including phases.
Mn3+, an ion of manganese(III), can function as an acid by giving a proton (H+) to a base. Here's an illustration: Mn3+ (aq) + 3OH- (aq) Mn(OH)3 (s)
What colour are Mn2+ and MnO4?There is no need to add an indicator because MnO4's vivid purple colour serves as one enough. In the conical flask, there is Fe2+. The Fe2+ solution is added, and the Fe2+ lowers the MnO4- to Mn2+. As Mn2+ is a colourless solution, the purple colour disappears.
What is the ion Mn2name? +'sThe divalent metal cation manganese(2+) contains manganese as the metal. It plays the part of a cofactor. It consists of a monoatomic dication, a manganese cation, and a divalent metal cation.
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Which of the following is the major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone?
Imine
The major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone is an imine.
A functional group or organic substance with a carbon-nitrogen double bond (C=N) is known as an imine. A hydrogen atom or an organic group may be joined to the nitrogen atom. (R). The carbon atom is connected to two more single bonds. Imines are present in numerous processes and are frequently found in manufactured and naturally occurring chemicals.
The five core atoms for ketimines and aldimines, C2C=NX and C(H)C=NX, respectively, are coplanar. The sp2-hybridization of the mutually double-bonded nitrogen and carbon atoms yields planarity. For nonconjugated imines, the C=N distance is 1.29-1.31, whereas for conjugated imines, it is 1.35. The C-N distances in amines and nitriles, on the other hand, are 1.47 and 1.16, respectively. Slow rotation occurs around the C=N bond. E- and Z-isomers were detected using NMR spectroscopy of aldimines have been detected. Owing to steric effects, the E isomer is favored.
An imine is formed when a primary amine reacts with a carbonyl group (C=O) of an aldehyde or ketone to form a new C-N bond. This reaction is known as a condensation reaction, as it involves the loss of a small molecule (e.g. water) to form the product.
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The correct questions is :
What is the major organic product of the condensation of ammonia or a primary amine with the carbonyl group of an aldehyde or ketone?
A 106 mL solution of a dilute acid is added to 157 mL of a base solution in a coffee-cup calorimeter. The temperature of the solution increases from 22.94 oC to 27.29 oC. Assuming the mixture has the same specific heat (4.184J/goC) and density (1.00 g/cm3) as water, calculate the heat (in J) transferred to the surroundings, qsurr.
Answer:
4897 J
Explanation:
The heat transferred to the surroundings, q_surr, can be calculated using the equation:
q_surr = -q_rxn = -CmΔT
where C is the specific heat capacity of the mixture (assumed to be the same as water, 4.184 J/g°C), m is the mass of the mixture (which we can calculate using the density, assuming that the volumes are additive), and ΔT is the change in temperature (in Celsius).
First, let's calculate the mass of the mixture:
density of water = 1.00 g/cm^3
volume of mixture = volume of acid + volume of base = 106 mL + 157 mL = 263 mL = 0.263 L
mass of mixture = density of water x volume of mixture = 1.00 g/cm^3 x 0.263 L = 263 g
Next, let's calculate the change in temperature:
ΔT = final temperature - initial temperature = 27.29°C - 22.94°C = 4.35°C
Now we can calculate the heat transferred to the surroundings:
q_surr = -CmΔT
q_surr = -(4.184 J/g°C) x (263 g) x (4.35°C)
q_surr = -4897 J
Note that the negative sign indicates that heat is lost by the system to the surroundings. Therefore, the heat transferred to the surroundings, q_surr, is 4897 J.
which the following optically active alcohol is treated with hbr, a racemic mixture of alkyl bromides is obtained
(S)-2-butanol will undergo an SN2 reaction with HBr to produce a racemic mixture of alkyl bromides. Here option B is the correct answer.
When optically active alcohol is treated with HBr, the reaction follows an SN1 or SN2 mechanism. In the case of SN1, a carbocation intermediate is formed, and in SN2, a backside attack by the nucleophile occurs. The stereochemistry of the product depends on the configuration of the intermediate and the direction of attack.
In the case of (S)-2-butanol, the hydroxyl group is attached to the second carbon atom, which makes it a primary alcohol. When treated with HBr, it undergoes an SN2 reaction, where the hydroxyl group is replaced by the bromine atom. The nucleophile attacks from the backside of the molecule, leading to an inversion of configuration.
This results in the formation of a racemic mixture of alkyl bromides, as both enantiomers have an equal chance of being attacked from either side. On the other hand, (R)-2-butanol, being the enantiomer of (S)-2-butanol, will also undergo the same reaction and produce the same racemic mixture of alkyl bromides.
In the case of (R)-1-phenyl ethanol and (S)-1-phenyl ethanol, they are secondary alcohols and can undergo either SN1 or SN2 reactions depending on the reaction conditions. However, the reaction mechanism will lead to the formation of a mixture of diastereomers, rather than a racemic mixture of enantiomers.
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Complete question:
Which of the following optically active alcohols, when treated with HBr, results in a racemic mixture of alkyl bromides?
a) (R)-2-butanol
b) (S)-2-butanol
c) (R)-1-phenyl ethanol
d) (S)-1-phenyl ethanol
Menthol is composed of C, H, and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g H2O. What is the empirical and molecular formula for menthol?
The empirical formula, CH2O9(menthol) is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.
The molecular formula for menthol is C5H10O.
This can be determined by dividing the molar mass of the empirical formula (156.26 g/mol) by the molar mass of CO2 (44.01 g/mol). This gives a ratio of 3.55, which is equal to the ratio of C atoms in the empirical formula, C10H20O.
Therefore, the molecular formula is C5H10O.
Given:
Menthol is composed of C, H, and O0.1005g sample of menthol is combusted and produces0.2829g of CO2 0.1159g H2O
1. Find: Empirical and molecular formula for menthol.
Let's first calculate the number of moles of CO2 produced. The balanced equation for combustion of menthol is:
C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)
From the above equation, we can see that for 10 moles of CO2 produced 1 mole of menthol is required.
2. By taking the number of moles of CO2 produced, we can calculate the number of moles of menthol burned.
Moles of CO2 = 0.2829g / 44.01g/mol= 0.00643 mol
C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)
From the balanced equation,1 mole of C10H20O requires 10 moles of CO2
Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol
Next, we can calculate the number of moles of H2O produced.
Moles of H2O = 0.1159g / 18.015g/mol= 0.00643 mol
C10H20O(s) + 12O2(g) → 10CO2(g) + 10H2O(l)
From the balanced equation,1 mole of C10H20O requires 10 moles of H2O
Moles of C10H20O burned = 10 * 0.00643= 0.0643 mol
3. Now we can calculate the empirical formula of menthol. The empirical formula can be calculated as follows:
Empirical formula = CH2O (Divide all moles by smallest moles) The molecular weight of CH2O = 30 g/mol
The empirical formula mass of the compound is:
mass = (12.011 + 2*1.008 + 15.999) = 30.026
Empirical formula mass of CH2O is 30.026g/mol, and the given sample weighs 0.1005 g.
The number of empirical formula units in the sample is 0.1005 g / 30.026 g/mol = 0.003348Units.
Empirical formula = CH2OThe empirical formula weight of menthol is CH2O, which is equal to 30.026g/mol.
4. To find the molecular formula, we need to know the molecular weight of the menthol. We can calculate it as follows:
Molecular formula mass = Empirical formula mass x n
Where n = integer Molecular formula mass of menthol is 156 g/mol, and the empirical formula mass is 30.026 g/mol.
So, n = 156 g/mol ÷ 30.026 g/mol = 5.192
Thus the empirical formula, CH2O is multiplied by 5 to get the molecular formula, C10H20O.The empirical and molecular formula for menthol are CH2O and C10H20O, respectively.
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which of the following statements may be true regarding a biochemical oxidation-reduction (redox) reaction?
A few statements may be true regarding a biochemical oxidation-reduction (redox) reaction. The statements are as follows: A redox reaction occurs when there is a transfer of electrons between molecules or atoms.
The electron donor becomes oxidized, and the electron acceptor is reduced, causing a transfer of energy. A redox reaction produces ATP, which is the primary energy currency of the cell. Oxidation and reduction are complementary reactions that occur simultaneously in the same reaction, resulting in the release of energy. Redox reactions are vital in metabolic pathways, and the electron carriers NAD+ and FAD+ are essential in these reactions. Oxygen is frequently used as a final electron acceptor in redox reactions. Redox reactions can also occur in non-cellular environments, such as photosynthesis, respiration, and combustion. The significance of redox reactions is enormous, and they play an essential role in sustaining life on earth. They help in generating energy, breaking down complex molecules, synthesizing molecules, and many other cellular processes.
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How many calories are required to raise the temperature of a 35.0 g sample from 35 °C to 85 °C? The sample has a specific heat of 0.108 cal/g °C.
Answer:
First, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 85 °C - 35 °C
ΔT = 50 °C
Next, we can use the following formula to calculate the heat energy required:
Q = m·C·ΔT
where Q is the heat energy in calories, m is the mass of the sample in grams, C is the specific heat in cal/g °C, and ΔT is the change in temperature in °C.
Plugging in the given values, we get:
Q = 35.0 g · 0.108 cal/g °C · 50 °C
Q = 189.0 calories
Therefore, 189.0 calories are required to raise the temperature of the sample from 35 °C to 85 °C
1) what is the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29% oxygen? worksheet
The empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29% oxygen is CH2O.
To calculate this, you need to first convert the percentage composition into mass composition. This is done by multiplying the percentages by the molecular weight of each element.
Carbon: 65.5% x 12 g/mol = 0.786 g/mol
Hydrogen: 5.5% x 1 g/mol = 0.055 g/mol
Oxygen: 29% x 16 g/mol = 0.464 g/mol
Now that you have the mass composition, you can calculate the moles of each element by dividing the mass of each element by its molar mass.
Carbon: 0.786 g/mol / 12 g/mol = 0.065 mol
Hydrogen: 0.055 g/mol / 1 g/mol = 0.055 mol
Oxygen: 0.464 g/mol / 16 g/mol = 0.029 mol
Finally, divide each element's moles by the smallest moles to get the empirical formula.
Carbon: 0.065 mol / 0.029 mol = 2.24 = 2 mol
Hydrogen: 0.055 mol / 0.029 mol = 1.90 = 1 mol
Oxygen: 0.029 mol / 0.029 mol = 1.00 = 1 mol
Therefore, the empirical formula of the molecule is CH2O.
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Which transition metal can form both a high and low spin complex? Zn2+, Cu2+, Mn3+, Ti2+
Answer: Manganese
Explanation:
With titanium, it only has two d electrons, so it can't form different high and low spin complexes. It doesn't matter because it will never fill the higher-energy orbitals. The total spin state turns out to be +1 (two unpaired d electrons, no matter what). Therefore, manganese will form both a high and low spin complex.
write a balanced equation for the redox reaction between calcium metal and oxygen gas
a balanced equation for the redox reaction: 2 Ca(s) + O2(g) → 2 CaO(s)
What is a redox reaction?A redox reaction is a type of chemical reaction that involves the transfer of electrons between species. One species undergoes oxidation (loses electrons) while another species undergoes reduction (gains electrons).
Which species is being oxidized and which species is being reduced in the reaction between calcium metal and oxygen gas?In the reaction between calcium metal and oxygen gas, the calcium metal is being oxidized (loses electrons) and the oxygen gas is being reduced (gains electrons). This can be seen in the balanced equation where the calcium atoms go from having an oxidation state of 0 to +2, while the oxygen atoms go from having an oxidation state of 0 to -2.
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Consider the following silica gel TLC plate of compounds A, B, and C developed in hexanes:
Consider the following silica gel TLC plate of com
a) Determine the R f values of compounds A, B, and C run on a silica gel TLC plate using hexanes as the solvent
b) Which compound, A, B, or C, is the most polar?
c) What would you expect to happen to the R f values if you used acetone instead of hexanes as the eluting solvent? (Think polarity of solvents)
The R f values for compounds A, B, and C on a silica gel TLC plate developed in hexanes would be determined by measuring the distance each compound traveled compared to the distance the solvent traveled.
a) There is a 4 cm gap between the origin and the solvent front. The Rf value for spot A is[tex]\frac{1.5}{4}= 0.375[/tex], because it travelled 1.5 cm. Due to the 3.5 cm movement of Spot B, its Rf is[tex]\frac{3.5}{4} = 0.875[/tex]. Spot C shifted 3 cm, making its Rf [tex]\frac{3}{4} = 0.75[/tex].
b)Due to its shorter travel distance than the other two compounds, compound A is the most polar. Recall that polar substances adhere to the adsorbent more readily, move less, and have a lower Rf value.
c)Hexanes is less polar than acetone as a solvent. Each of the three compounds would move more quickly if the same method were employed to elute them.The chemicals can be removed from the polar adsorbent more effectively with a more polar eluting solvent. Each compound would have a higher Rf value if acetone were used to elute the TLC plate as opposed to hexanes because each compound travels more quickly.
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What is the amount of pi?
However, it is commonly approximated as 3.14159.
What is an irrational number ?An irrational number is a number that cannot be expressed as a simple fraction or ratio of two integers. It is a non-repeating, non-terminating decimal. Examples of irrational numbers include pi (π), the square root of 2 (√2), and the golden ratio (∅).
What is a termination ?In mathematics, a terminating decimal is a decimal number that has a finite number of digits after the decimal point, i.e., the decimal representation ends in a finite number of zeroes. For example, 0.75, 2.0, and 0.0625 are terminating decimals.
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How many atoms of lithium are in 18.7 g?
The atoms of lithium that are in 18.7 g is 16 × 10²³ atoms . This is taken out by mole concept .
What is mole concept ?The mole is a unit of measurement similar to the pair, dozen, gross, and so on. It provides a precise count of the atoms or molecules in a bulk sample of matter. A mole is the amount of substance that contains the same number of discrete entities (atoms, molecules, ions, etc.)
if 7 grams of lithium contain 6 × 10²³ atoms
then 18.7 will contain 16 × 10²³ atoms
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The titration of 45.0 ml of an unknown triprotic acid required 32.71 ml of 0.37 M KOH to
reach the endpoint. What is the molarity of the unknown acid?
The molarity of the unknown triprotic acid is 0.269M.
How to calculate molarity?Molarity is the concentration of a substance in solution, expressed as the number moles of solute per litre of solution.
The molarity of the unknown acid can be calculated using the following formula:
CaVa = CbVb
Where;
Ca and Va = acid concentration and volume respectivelyCb and Vb = base concentration and volume respectivelyAccording to this question, the titration of 45.0 ml of an unknown triprotic acid required 32.71 ml of 0.37 M KOH to reach the endpoint.
45 × Ca = 32.71 × 0.37
45Ca = 12.1027
Ca = 0.269M
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What is the pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI? The Kb of trimethylamine, (CH3)3N, is 6.3 x 10-5. (value = 0.02)
The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.
What is pH?
To find the pH of the solution, we need to first determine if (CH3)3NHCI acts as an acid or base. Since (CH3)3NHCI is a salt composed of a weak base, trimethylamine, and a strong acid, hydrochloric acid (HCI), it will undergo hydrolysis in water.
The hydrolysis reaction is given by:
(CH3)3NH+ (aq) + H2O (l) ⇌ (CH3)3N (aq) + H3O+ (aq)
The Kb expression for the equilibrium reaction is:
Kb = [ (CH3)3N ] [ H3O+ ] / [ (CH3)3NH+ ]
Since (CH3)3NH+ and HCl dissociate completely in water, the initial concentration of (CH3)3NH+ is equal to the concentration of (CH3)3NHCI, which is 0.335 M.
Using the Kb value given, we can solve for the concentration of H3O+:
Kb = 6.3 x [tex]10^{-5}[/tex] = [ (CH3)3N ] [ H3O+ ] / 0.335
[ H3O+ ] = Kb x (CH3)3NH+ / (CH3)3N
[ H3O+ ] = 6.3 x [tex]10^{-5}[/tex] x 0.335 / 1
[ H3O+ ] = 2.1095 x [tex]10^{-6}[/tex] M
Finally, we can calculate the pH using the expression:
pH = -log [H3O+]
pH = -log (2.1095 x [tex]10^{-6}[/tex])
pH = 5.676
Therefore, the pH of the solution is approximately 5.676.
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Complete question is: The pH of 0.335 M trimethylammonium chloride, (CH3)3NHCI is approximately 5.676.
!!!50 points!!!
Problem 1. What masses of 15% and 20% solutions are needed to prepare 200 g of 17% solution?
Problem 2. What masses of 18% and 5% solutions are needed to prepare 300 g of 7% solution?
Problem 3. 200 g of 15% and 350 g of 20% solutions were mixed. Calculate mass percentage of final solution.
Problem 4. 300 g of 15% solution and 35 g of solute were mixed. Calculate mass percentage of final solution.
Problem 5. 400 g of 25% solution and 150 g of water were mixed. Calculate mass percentage of final solution.
Answer:
See Below.
Explanation:
Problem 1
Let x be the mass of 15% solution needed and y be the mass of 20% solution needed. Then, we have the following system of equations:
x + y = 200 (total mass of solution)
0.15x + 0.20y = 0.17(200) (total amount of solute)
Solving this system of equations gives:
x = 60 g (mass of 15% solution)
y = 140 g (mass of 20% solution)
Therefore, 60 g of 15% solution and 140 g of 20% solution are needed to prepare 200 g of 17% solution.
Problem 2
Let x be the mass of 18% solution needed and y be the mass of 5% solution needed. Then, we have the following system of equations:
x + y = 300 (total mass of solution)
0.18x + 0.05y = 0.07(300) (total amount of solute)
Solving this system of equations gives:
x = 120 g (mass of 18% solution)
y = 180 g (mass of 5% solution)
Therefore, 120 g of 18% solution and 180 g of 5% solution are needed to prepare 300 g of 7% solution.
Problem 3
The total mass of the final solution is
200 g + 350 g = 550 g
The total amount of solute in the final solution is:
0.15(200 g) + 0.20(350 g) = 95 g + 70 g = 165 g
Therefore, the mass percentage of the final solution is:
(mass of solute / total mass of solution) x 100% = (165 g / 550 g) x 100% = 30%
Therefore, the mass percentage of the final solution is 30%.
Problem 4
The total mass of the final solution is
300 g + 35 g = 335 g
The total amount of solute in the final solution is:
0.15(300 g) + 35 g = 75 g + 35 g = 110 g
Therefore, the mass percentage of the final solution is:
(mass of solute / total mass of solution) x 100% = (110 g / 335 g) x 100% = 32.8%
Therefore, the mass percentage of the final solution is 32.8%.
Problem 5
The total mass of the final solution is
400 g + 150 g = 550 g
The total amount of solute in the final solution is
0.25(400 g) = 100 g
Therefore, the mass percentage of the final solution is
(mass of solute / total mass of solution) x 100% = (100 g / 550 g) x 100% = 18.2%
Therefore, the mass percentage of the final solution is 18.2%.
Which statement below correctly describes their relative atomic radii and first ionization energy when comparing Se and Br? The atomic radius for Se is larger than Br, and the first ionization energy for Se is greater than Br. The atomic radius for Br is larger than Se, and the first ionization energy for Bris greater than Se. The atomic radius for Se is larger than Br, and the first ionization energy for Br is greater than Se. The atomic radius for Br is larger than Se, and the first ionization energy for Se is greater than Br.
At has a higher initial ionisation energy than Br, while Br has a bigger atomic radius. Se has a bigger atomic radius than Br, and Br has a higher initial ionisation energy than Se.
How do atomic radii and ionisation energy relate to one another (i.e., what happens to ionisation energy as atomic radii grow)?The most loosely bound electron is further from the nucleus and thus easier to remove in bigger atoms. Hence, the ionisation energy should decrease as size (atomic radius) increases.
Why does ionisation energy rise across a period while decreasing down a group?This is because the outer electrons aren't bound as strongly because they are farther from the nucleus.
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A student sets up a titration with a * 1 point buret filled with 0.5 M NaOH. In the flask below they place the phenolphthalein indicator and 6.2 mL of the unknown acid. The solution in the beaker turns pink after exactly 24.8 mL of NaOH have been added. Find the exact concentration of the unknown acid.
2. Hydrogen bromide reacts with propene to form either 1-bromopropane or 2-bromopropane. Explain why
2-bromopropane is the major product.
3. Explain how the reaction with bromine can be used to test for an alkene. Include the mechanism for the reaction between hex-1-ene and bromine in your answer.
a) Describe the process of addition polymerisation.
b) Show the repeating unit of the polymer that is formed from the addition polymerisation of chloroethene monomers. Name and give at least one use for this polymer.
Answer:
Explanation:
2-bromopropane is the major product because the reaction mechanism involves the formation of the most stable carbocation intermediate. When hydrogen bromide reacts with propene, the hydrogen atom from HBr adds to the carbon atom of the double bond that has fewer hydrogen atoms attached, resulting in the formation of a carbocation intermediate. The intermediate can either form 1-bromopropane or 2-bromopropane depending on the position of the carbocation. The 2-bromopropane is the major product because the secondary carbocation formed in this case is more stable than the primary carbocation formed in the case of 1-bromopropane.
To test for an alkene, bromine water can be used. When an alkene reacts with bromine water, the bromine molecule adds across the double bond, forming a colorless dibromoalkane product. The mechanism for the reaction between hex-1-ene and bromine involves the formation of a cyclic bromonium ion intermediate, followed by the attack of water on the intermediate, resulting in the formation of the dibromoalkane product.
a) Addition polymerization is a process in which unsaturated monomers are joined together to form a polymer. The process involves breaking the double bond of the monomer and joining the monomers together to form a long-chain polymer. The process requires a catalyst to initiate the reaction.
b) The repeating unit of the polymer formed from the addition polymerization of chloroethene monomers is -CH2-CHCl-. This polymer is called polyvinyl chloride (PVC), and it has a wide range of uses, including pipes, electrical cables, and vinyl flooring.
what happens when zinc chloride reacts with potassium hydroxide and what formed?
Answer:
when the solution of potassium hydroxide and zinc chloride are mixed,the double-displacement reaction occur ,resulting in precipitation and the reaction forms potassium chloride and zinc hydroxide .
A 0.036 M aqueous nitrous acid (HNO2) solution has an osmotic pressure of 0.93 atm at 25°C. Calculate the percent ionization of the acid.
The percent ionization of the nitrous acid in the 0.036 M aqueous solution is 2.1%.
How to calculate the percent ionization of the acid ?
The osmotic pressure (π) of a solution can be related to the molar concentration (M) of the solute and the temperature (T) of the solution by the following equation:
π = MRT
Where R is the gas constant.
We can use this equation to calculate the molar concentration of the nitrous acid solution:
M = π / RT
M = (0.93 atm) / (0.0821 L·atm/(mol·K) x 298 K)
M = 0.036 M
This is the molar concentration of the undissociated nitrous acid in the solution. To calculate the percent ionization of the acid, we need to know the concentration of the H+ and NO2- ions in the solution.
The balanced chemical equation for the dissociation of nitrous acid is:
HNO2(aq) ⇌ H+(aq) + NO2-(aq)
Let x be the extent of ionization of the nitrous acid. Then the concentration of H+ and NO2- ions can be expressed in terms of x as follows:
[H+] = x M
[NO2-] = x M
The concentration of the undissociated nitrous acid is (1-x)M.
The expression for the equilibrium constant (Ka) of the reaction can be written as:
Ka = [H+] [NO2-] / [HNO2]
Substituting the concentrations in terms of x, we get:
Ka = x^2M / (1-x)M
Simplifying the above equation, we get:
Ka = x^2 / (1-x)
The percent ionization of the acid is the fraction of the original HNO2 molecules that dissociate into H+ and NO2- ions. It can be calculated as follows:
% ionization = (concentration of H+ ions) / (initial concentration of HNO2) x 100
% ionization = (x M) / (M) x 100
% ionization = x x 100
Substituting the value of x from the above equation for Ka, we get:
Ka = x^2 / (1-x)
x = sqrt(Ka / (1+Ka))
We can calculate the value of Ka using the standard reference value of the acid dissociation constant (Ka) for nitrous acid at 25°C, which is 4.5 x 10^-4.
x = sqrt(4.5 x 10^-4 / (1+4.5 x 10^-4))
x = 0.021
% ionization = 0.021 x 100
% ionization = 2.1%
Therefore, the percent ionization of the nitrous acid in the 0.036 M aqueous solution is 2.1%.
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Chemistry Help Please! It's worth a lot of points
1.Write the equilibrium expression for the following reactions
a. H2SO4(aq) + H2O(L) ⇆ HSO4-(aq) + H3O+(aq)
b. 4NH3(g) + 5O2(g) ⇆ 4NO(g) + 6H2O(g)
c. NH4Cl(s) ⇆ NH3(g) + HCl(g)
d. N2O4(g) ⇆ 2NO2(g)
2. The following reaction has a K value of 0.050. What does that mean about the concentrations of the reactants as compared to the products? Be specific in your answer.
N2(g) + 3H2(g) ⇆ 2NH3(g)
3. The following reaction has a K value of 6.8 x 103. What does that mean about the concentrations of the reactants as compared to the products? Be specific in your answer.
2SO3(g) ⇆ 2SO2(g) + O2(g)
4. When dissolving substances in water, the degree of solubility of a substance is often represented as the solubility product constant (Ksp). The solubility product constant is the same thing as the equilibrium constant for the dissolving reaction. Two substances that dissociate in water are shown below alone with the Ksp.
NaCl(s) ⇆ Na+(aq) + Cl-(aq) Ksp = 36
BaSO4(s) ⇆ Ba2+(aq) + SO42-(aq) Ksp = 1.1 x 10-16
5. Identify and label the Brønsted-Lowry acid, its conjugate base, the Brønsted-Lowry base, and its conjugate acid in each of the following equations:
a. HNO3 + H2O ⟶ H3O+ + NO3−
b. CN− + H2O ⟶ HCN + OH−
c. H2SO4 + Cl− ⟶ HCl + HSO4−
d. HSO4− + OH− ⟶ SO42− + H2O
e. O2− + H2O ⟶2OH−
6. What is the conjugate acid of each of the following? What is the conjugate base of each of the following?
a. OH-
b. H2O
c. HCO3-
d. NH3
e. HSO4-
7. The following acids are shown with their equilibrium constants (also known as the acid dissociation constant). Rank these acids from strongest to weakest. Explain your ranking.
HCN(aq) + H2O(L) ⇆ H3O+(aq) + CN-(aq) K = 6.2 x 10-10
HC2H3O2(aq) + H2O(L) ⇆ H3O+(aq) + C2H3O-(aq) K = 1.75 x 10-5
H2CO3(aq) + H2O(L) ⇆ H3O+(aq) + HCO3-(aq) K = 4.5 x 10-7
HIO4(aq) + H2O(L) ⇆ H3O+(aq) + IO4-(aq) K = 2.3 x 10-2
8. Calculate the pH and the pOH of each of the following solutions.
a. 0.200 M HCl
b. 0.0143 M NaOH
c. 3.0 M HNO3
d. 0.0031 M Ca(OH)2
9. Wine has a pH of 3.6. What are the hydronium and hydroxide ion concentrations?
10. The hydroxide ion concentration in household ammonia is 3.2 x 10-3 M. What is the concentration of hydronium ions?
Answer:
1. Equilibrium expressions:
a. K = [HSO4-][H3O+]/[H2SO4][H2O]
b. K = [NO]^4[H2O]^6/[NH3]^4[O2]^5
c. K = [NH3][HCl]/[NH4Cl]
d. K = [NO2]^2/[N2O4]
2. Since K = 0.050, the concentrations of the reactants (N2 and H2) are larger than the concentrations of the products (NH3).
3. Since K = 6.8 x 10^3, the concentrations of the products (SO2 and O2) are larger than the concentrations of the reactant (SO3).
4. The Ksp expression for each of the reactions is:
a. Ksp = [Na+][Cl-]
b. Ksp = [Ba2+][SO42-]
5. Brønsted-Lowry acids and bases:
a. Acid: HNO3; Conjugate base: NO3-; Base: H2O; Conjugate acid: H3O+
b. Acid: HCN; Conjugate base: CN-; Base: H2O; Conjugate acid: HCN
c. Acid: H2SO4; Conjugate base: HSO4-; Base: Cl-; Conjugate acid: HCl
d. Acid: NH3; Conjugate base: NH2-; Base: H2O; Conjugate acid: NH4+
e. Acid: H2O; Conjugate base: OH-; Base: O2-; Conjugate acid: OH-
6. Conjugate acids and bases:
a. Acid: H2O; Conjugate base: OH-
b. Acid: H3O+; Conjugate base: H2O
c. Acid: H2CO3; Conjugate base: HCO3-
d. Acid: NH4+; Conjugate base: NH3
e. Acid: HSO4-; Conjugate base: SO42-
7. The strongest acid is HIO4 (highest K value), followed by HCN, HC2H3O2, and H2CO3 (lowest K value). The K values represent the degree to which the acids dissociate in solution. HIO4 is a strong acid, meaning it dissociates almost completely in solution, while H2CO3 is a weak acid, meaning it only dissociates partially.
8. pH and pOH calculations:
a. pH = -log[H3O+] = -log(0.200) = 0.699; pOH = -log[OH-] = -log(1.0 x 10^-14/0.200) = 12.301
b. pOH = -log[OH-] = -log(0.0143) = 1.844; pH = 14.000 - pOH = 12.156
c. pH = -log[H3O+] = -log(3.0) = 0.522; pOH = 13.478
d. pOH = -log[OH-] = -log(0.0062) = 2.206; pH = 14.000 - pOH = 11.794
9. Hydronium and hydroxide ion concentrations:
pH = 3.6; hydronium ion concentration = 10^-pH = 3.98 x 10^-4 M; hydro
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FILL IN THE BLANK. Use the Gizmo to find the freezing, melting, and boiling points of water at 5,000 meters (16,404 feet). Write these values below. Freezing point: _______ Melting point: _______ Boiling point: _______
If we use the Gizmo to find the freezing, melting, and boiling points of water at 5,000 meters (16,404 feet) then,
Freezing point: 32 ºF (0ºC)
Melting point: 32 ºF (0ºC)
Boiling point: 203°F (95°C)
The freezing point is defined as the temperature at which a liquid becomes a solid. Increased pressure usually raises the freezing point with the melting point of the solid. The boiling point of a pure substance is defined as the temperature at which the substance transitions from a liquid to the gaseous phase. At the boiling point the vapor pressure of the liquid is equal to the applied pressure on the liquid. The melting point of a substance is defined as the temperature at which the substance changes from a solid to a liquid.
Melting occurs at a single temperature for the pure substances. The normal and average melting point and boiling point of water at 1 atmospheric pressure are 0°C and 100°C respectively. Decreasing the pressure under 1 atm. will lower the boiling point since the external pressure will be lower so it will become equal with the vapor pressure at a lower temperature.
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What is the bond angle of carbonothioyl dibromide
Also what is the molecular shape
Answer:
Carbonothioyl dibromide, also known as CBr2S, has a bond angle of approximately 109.5 degrees, which is the typical tetrahedral bond angle for molecules with sp3 hybridization.
The molecular shape of CBr2S is also tetrahedral, with the two bromine atoms and the sulfur atom arranged at the corners of a tetrahedron, and the carbon atom at the center.
