Answer:
The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].
The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].
Explanation:
The oxidation states of atoms in a compound should add up to zero.
Ag₂OThere are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:
[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].
SO₂Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.
Therefore:
[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the oxidation state of [tex]\rm S[/tex] here:
[tex]\text{Oxidation state of $\rm S$} = 4[/tex].
Calculate the molarity of bromide ions in 250. mL of a solution containing 25.9 g NaBr and 0.155 moles of HBr.
Answer:
[tex]1.628 M[/tex]
Explanation:
From the question we were given 0.155 moles of HBr, but Br and H are in ratio 1:1, then there are 0.155 moles of Br- ions.
We were also told that the solution contain NaBr, of 25.9 g. Then it must be converted to moles.
molar mass of NaBr =(22.99g + 79.90 )
= 102.89 g per mol.
the moles of NaBr can be calculated as 25.9 / 102.89
=0.252 moles
But Na and Br are in a ratio 1:1 , then there are 0.252 moles of Br-.
Then to get two Br- mol , we will add the first and second mol of Br- together
= 0.155 + 0.252
=0.407 moles.
The given solution has volume of 250 mL, but we know that there are 1000 ml in a liter, then if we convert to L for unit consistency we have
= 250/1000
= 0.25 L
molarity=0.407 moles/0.25 L
= 1.628 M.
Therefore, Br ion molarity is 1.628 M.
The molarity of the Br ions in the 250 ml solution has been 1.628 M.
Moles can be defined as the mass per unit molecular mass. Moles can be expressed as:
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of NaBr = [tex]\rm \dfrac{25.9}{102.89}[/tex]
Moles of NaBr = 0.252 mol
Moles of HBr = 0.155 mol.
Since both the compounds have 1:1 ratio of atom: Br, the Br produced has been equal to the concentration of the compound.
Br from NaBr = 0.252 mol
Br from HBr = 0.155 mol.
Total Br ions = 0.407 mol.
Molarity can be expressed as:
Molarity = [tex]\rm moles\;\times\;\dfrac{1000}{Volume\;(ml)}[/tex]
Molarity of Br ions = 0.407 × [tex]\rm \dfrac{1000}{250\;ml}[/tex]
Molarity of Br ions = 1.628 M.
The molarity of the Br ions in the 250 ml solution has been 1.628 M.
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Zeros laced at the end of the significant number are...
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
13. If atoms from two different elements react to form a compound, the element with a higher
number
will have a negative oxidation
A. atomic radius
B. energy
C. electronegativity
D. atomic number
Answer:
C. electronegativity
Explanation:
THE ELEMENT WITH THE GREATE ELECTRONEGATIVITY IS ASSIGNED A NEGATIVE OXIDATION NUMBER EQUAL TIO ITS CHARGE
The standard free energy change for a reaction can be calculated using the equation ΔG∘′=−nFΔE∘′ ΔG∘′=−nFΔE∘′ where nn is the number of electrons transferred, FF is Faraday's constant, 96.5 kJ·mol−1·V−1, and ΔE∘′ΔE∘′ is the difference in reduction potential. For each of the given reactions, determine the number of electrons transferred (n)(n) and calculate standard free energy (ΔG∘′)(ΔG∘′) . Consider the half-reactions and overall reaction for reaction 1. half-reactions:fumarate 2−+2H+CoQH2↽−−⇀succinate−↽−−⇀CoQ+2H+ half-reactions:fumarate−+2H+↽−−⇀succinate2−CoQH2↽−−⇀CoQ+2H+ overall reaction:fumarate2−+CoQH2↽−−⇀succinate2−+CoQΔE∘′=−0.009 V
Answer:
ΔG°′ = 1.737 KJ/mol
Explanation:
The reaction involves the transfer of two electrons in the form of hydride ions from reduced coenzyme Q, CoQH₂ to fumarae to form succinate and oxidized coenzyme Q, CoQ.
The overall equation of reaction is as follows:
fumarate²⁻ + CoQH₂ ↽⇀ succinate²⁻ + CoQ ; ΔE∘′=−0.009 V
Using the equation for standard free energy change; ΔG°′ = −nFΔE°′
where n = 2; F = 96.5 KJ.V⁻¹.mol⁻¹; ΔE°′ = 0.009 V
ΔG°′ = - 2 * 96.5 KJ.V⁻¹.mol⁻¹ * 0.009 V
ΔG°′ = 1.737 KJ/mol
What is the concentration of MgSO4 in a solution prepared by dissolving 30g MgSO4 in 500ml distilled water. Express concentration in
(i)ppm
(ii) %w/v
(iii) %w/w
Assume the solution density is 1.15g/ml.
Answer:
Concentration of MgSO4 = 0.0521 × 10⁶ ppmConcentration of MgSO4 = 6% w/vConcentration of MgSO4 = 5.21% w/wExplanation:
Given:
Mass of solute = 30 gram
Volume of water = 500 ml
Density = 1.15g/ml
Find:
(i)ppm
(ii) %w/v
(iii) %w/w
Computation:
Water in gram = 500 ml × 1.15 g/ml
Water in gram = 575 gram
In ppm
Concentration of MgSO4 = [30 / 575] × 10⁶
Concentration of MgSO4 = 0.0521 × 10⁶ ppm
in % w/v
Concentration of MgSO4 = [30 / 500] × 100
Concentration of MgSO4 = 6% w/v
in % w/w
Concentration of MgSO4 = [30 / 575] × 100
Concentration of MgSO4 = 5.21% w/w
During a titration, a known concentration of _____ is added to a _____ of an unknown concentration g
Explanation:
The whole process of titration involves finding the concentration of a solution (usually an acid or base) by adding (titrating) it to a solution(acid or base) with a known concentration.
The solution of unknown concentration (the analyte) is usually placed in an flask, while the solution of known concentration (titrant) is placed in a burette and slowly added to the flask.
Which of the following is a salt that will form from the combination of a strong base with a weak acid?
Select the correct answer below:
A. NaHCO3
B. H2O
C. CH3CO2H
D. NH4Cl
Answer:
A. NaHCO₃
Explanation:
NaHCO₃ ⇒ NaOH + H₂CO₃
NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.
Hope that helps.
From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
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A sample of pure lithium chloride contains 16% lithium by mass. What is the % lithium by mass in a sample of pure lithium carbonate that has twice the mass of the first sample
Answer:
Percentage lithium by mass in Lithium carbonate sample = 19.0%
Explanation:
Atomic mass of lithium = 7.0 g; atomic mass of Chlorine = 35.5 g; atomic mass of carbon = 12.0 g; atomic mass of oxygen = 16.0 g
Molar mass of lithium chloride, LiCl = 7 + 35.5 = 42.5 g
Percentage by mass of lithium in LiCl = (7/42.5) * 100% = 16.4 % aproximately 16%
Molar mass of lithium carbonate, Li₂CO₃ = 7 * 2 + 12 + 16 * 3 =74.0 g
Percentage by mass of lithium in Li₂CO₃ = (14/74) * 100% = 18.9 % approximately 19%
Mass of Lithium carbonate sample = 2 * 42.5 = 85.0 g
mass of lithium in 85.0 g Li₂CO₃ = 19% * 85.0 g = 16.15 g
Percentage by mass of lithium in 85.0 g Li₂CO₃ = (16.15/85.0) * 100 % = 19.0%
Percentage lithium by mass in Lithium carbonate sample = 19.0%
By December 31, 2003, concerns over arsenic contamination had prompted the manufacturers of pressure-treated lumber to voluntarily cease producing lumber treated with CCA (chromated copper arsenate) for residential use. CCA-treated lumber has a light greenish color and was widely used to build decks, sand boxes, and playground structures.
Required:
Draw the Lewis structure of the arsenate ion (ASO4^3-) that yields the most favorable formal charges.
Answer:
Explanation:
lewis structure can be defined as a process of how the valence shell electrons of a molecule is being arranged, the pattern of it arrangement and the relationship between the bonding atoms and the lone pairs present in the molecule.
In order to draw the Lewis structure for Arsenate ion [tex]\mathsf{AsO \ _4^{3-}}[/tex], first thing is to count the valence electrons in the molecule. Once we determine the valence electrons, then we distribute them around the central atom. The Arsenate ion structure is tetrahedral in nature with a bond angle of 109.5° and it is sp³ hybridized.
What's the mass in grams of 0.442 moles of calcium bromide, CaBr2? The atomic
weight of Ca is 40.1 and the atomic weight of Br is 79.9.
A) 452.3 g
B) 53.04 g
C) 44.2 g
D) 88.4 g
Answer:
Below
Explanation:
Let n be the quantity of matter in the Calcium Bromide
● n = m/ M
M is the atomic weight and m is the mass
M of CaBr2 is the sum of the atomic wieght of its components (2 Bromes atoms and 1 calcium atom)
M = 40.1 + 2×79.9
● 0.422 = m/ (40.1+2×79.9)
●0.422 = m/ 199.9
● m = 0.422 × 199.9
● m = 84.35 g wich is 88.4 g approximatively
88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 ,therefore option (d) is correct.
What do you mean by mass ?Mass is the amount of matter that a body possesses. Mass is usually measured in grams (g) or kilograms (kg) .
To calculate mass in grams of 0.442 moles of calcium bromide, CaBr2,
Let n be the quantity of matter in the Calcium Bromide
M is the atomic weight and m is the mass
n = m/ MM of CaBr2 is the sum of the atomic weight of its components
Mass of Ca = 40.1 , Mass of Br = 79.9
M = 40.1 + 2×79.9
0.422 = m/ (40.1+2×79.9)
0.422 = m/ 199.9
m = 0.422 × 199.9
m = 84.35 g which is 88.4 g approximatively .
Thus ,88.4 g approximatively is the mass in grams of 0.442 moles of calcium bromide, CaBr2 , hence option (d) is correct .
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A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
g If the titration of a 10.0-mL sample of sulfuric acid requires 28.15 mL of 0.100 M sodium hydroxide, what is the molarity of the acid
Answer:
[tex]M_{acid}=0.141M[/tex]
Explanation:
Hello,
In this case, the reaction between sulfuric acid and hydroxide is:
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
We can notice a 1:2 molar ratio between the acid and the base respectively, therefore, at the equivalence point we have:
[tex]2*n_{acid}=n_{base}[/tex]
And in terms of volumes and concentrations:
[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]
So we compute the molarity of sulfuric acid as shown below:
[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} =\frac{0.100M*28.15mL}{2*10.0mL}\\ \\M_{acid}=0.141M[/tex]
Best regards.
Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
ΔHrxn = -635.14kJ/molConsider this synthesis of isoamyl acetate based on this week's experimental methods, after refluxing the reaction mixture for 25 minutes, what is likely present in solution
Answer:
acetic acid and phosphoric acid
Explanation:
After refluxing the reaction mixture ( synthesis of isoamyl acetate ) what is likely present in the solution is acetic acid and phosphoric acid, this due to the fact that if the reaction time between the reactants was less than the refluxing time which is 25 minutes,
there will be no reactant ( 3-methylbutanol )left in the reaction mixture
A 50.0 L cylinder of oxygen gas is stored at 150. atm. What volume would the oxygen gas occupy if the cylinder were opened into a hot air balloon (completely deflated) until the final pressure is 735 torr
Answer:
THE VOLUME OF THE OXYGEN GAS AFTER DEFLATION TILL A PRESSURE OF 735 TORR IS ATTAINED IS 7836.99 L
Explanation:
Using Boyle's law,
P1V1 = P2V2
P1 = 150 atm
V1 = 50 L
P2 = 735 Torr
V2 = unknown
We must first convert the pressures into the same SI unit for easy calculation
1torr = 1/760 atm
So converting 735 torr to atm; we have:
1 torr = 1/ 760 atm
735 torr = 735 * 1 / 760 atm
= 0.967 atm
In other words, P2 = 0.957 atm
So rearranging the formula by making V2 the subject of the equation, we have:
V2 = P1 V1 / P2
V2 = 150 * 50 / 0.957
V2 = 7836.99 L
The volume of the oxygen cylinder after deflation to a final pressure of 735 torr or 0.967 atm pressure is 7836.99 L.
1. Why is it not possible to resolve the compound CH3-NH-CH2-CH3 into a pair of enantiomers?
2. Which one of the following is not affected (or is least affected) by the lone pair of electrons on an amine's nitrogen?
a. solubility in alcohols and in water.
b. hydrogen-bond formation.
c. melting point.
d. dipole moment.
e. basicity.
3. Which of the following compounds is most basic?
a. cyclohexyl amine.
b. p-nitroaniline.
c. 2,6-dimethylaniline.
d. p-methoxyaniline.
d. aniline.
Answer:
1. In the compound, H3C-NH-CH2-CH3, there are no chiral centers present, chiral centers refer to the configuration in which carbon is attached with four different groups. In the molecules, as there are no chiral centers, therefore the molecule is optically inactive, that is, it will not demonstrate pair of an enantiomer is one of the essential characteristics of optically active compounds is the possession of enantiomeric pairs.
2. On the nitrogen of aniline, the lone pair of electrons can produce hydrogen bonds, play an essential function in basicity, play an essential role in dipole moment or polarity, and wit the increase in solubility there is an increase in the formation of the hydrogen bond, eventually increasing to boiling point. However, the melting point is not affected. As the melting point is the characteristic of the packing efficacy of a molecule and does not rely upon the anilinic nitrogen's lone pairs.
3. With the increase in the tendency to donate an electron, basicity increases. However, if the electron is taking part in resonance, the donation will not take place easily, and the compound will be the least basic. Apart from cyclohexyl amine, in all the other given compounds, the lone pair of nitrogen takes part in the process of delocalization or conjugation. Thus, cyclohexyl amine will be most basic as the lone pairs are easily available for donation.
If sulfur gained another electron, would its charge be positive or negative?
Explain your thinking. *
Answer:
AS WE KNOW THAT , when non-metallic elements gain electrons to form anions, SO sulphur is non metal and have the capacity to gain two electrons as lies in 6th group so it can gain electron and become sulphide ion(S-).
Thanks for asking questionExplanation:
Q 13.3: Which of the following is the least stable radical choice? Tertiary radical. B : Allyl radical. C : Secondary radical. D : Methyl radical. E : Primary radical.
Answer:
Methyl radical
Explanation:
A radical is any specie that contains an odd number of electrons. We must note that the greater the number of alkyl groups which are attached to a carbon atom that bears the odd electrons, the more the degree of delocalization of the odd electrons and consequently the more stable we expect the free radical to be.
Hence the order of free radical stability is; Methyl < Primary < Secondary < Tertiary. Hence, we can easily see that the methyl radical is the least stable free radical.
Answer: Methyl radical
Explanation:
Im really confused and select all that apply questions scare me.
Answer:
The 3rd one
Explanation:
In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
Answer:
The volume of the container is 59.112 L
Explanation:
Given that,
Number of moles of Oxygen, n = 3
Temperature of the gas, T = 300 K
Pressure of the gas, P = 1.25 atm
We need to find the volume of the container. For a gas, we know that,
PV = nRT
V is volume
R is gas constant, R = 0.0821 atm-L/mol-K
So,
[tex]V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L[/tex]
So, the volume of the container is 59.112 L
Which of the following substances (along with its corresponding salt) would be best suited for generating a buffer solution with a pH below 7?
a. CH3CO2H
b. C5H5N
c. HCl
d. None of the above
Answer:
d. None of the above
Explanation:
A buffer works when pH of the buffer is ± 1. Out of this range, the buffering capacity is very low.
Acetic acid, CH₃CO₂H, has a pKa of 4.74. That means its buffering capacity is between 3.74 and 5.74 of pH. Is not a good buffer to pH 7
Pyridine is a weak base with pKa of 5.52. Its buffering capacity is between 6.52 and 4.52. Is not a good buffer to pH 7
HCl is a strong acid. Just weak acids and bases can produce a buffer with its conjugate base. HCl can't produce a buffer.
Thus, right answer is:
d. None of the aboveAnswer:
CH3CO2H
Explanation:
A weak acid, such as acetic acid, paired with its conjugate base (here, the acetate ion) will be an ideal system for creating an acidic buffer solution with a pH below 7.
2NH3 → N2 + 3H2 If 2.22 moles of ammonia (NH3) decomposes according to the reaction shown, how many moles of hydrogen (H2) are formed? A) 2.22 moles of H2 B) 1.11 moles of H2 C) 3.33 moles of H2 D) 6.66 moles of H2
Answer:
C
Explanation:
According to the mole ratio, using 2NH3 will give you 3H2. Which means in order to find the moles of H2 you would only need to divide 2 and multiply 3 to get the amount of moles of H2 produced.
Answer:
I think it's C
Explanation:
Please, tell me if I'm incorrect.
Al diluir 25 g de sal de mesa en 250ml de agua, ¿En cuántos °C aumenta el punto de ebullición de la disolución formada? ( Ke = 0,52 °C/molal , PM NaCl = 58,44 g/mol)
Answer:
ΔT=[tex]0.87^{\circ}C[/tex]
Explanation:
Para esta pregunta debemos recordar la ecuación que nos permite calcular el aumento ebulliscopico (aumento del punto de ebullición):
ΔT=[tex]Kb*m[/tex]
Donde ΔT es el valor del aumento del punto de ebullición. Kb es la constante ebulloscopica para el agua ([tex]0.512\frac{Kg~^{\circ}C}{mol}[/tex]) y m es la molalidad ([tex]m=\frac{mol}{Kg~ of~ solvente}[/tex]).
Por lo tanto el primer paso es calcular la molalidad de la solución. Para lo cual tendremos que calcular las moles de sal en los 25 g. Si queremos hacer esto debemos recordar que la formula de sal de mesa es NaCl y que la masa molar de NaCl es 58.44 g/mol. Por lo tanto:
[tex]25~g~NaCl\frac{1~mol~NaCl}{58.44~g~NaCl}~=~0.42~mol~NaCl[/tex]
Ahora bien, también debemos saber los Kg de agua en la solución. Por lo que podemos usar la densidad del agua (1 g/mL) para convertir de mL a g y luego hacer la conversión a Kg:
[tex]250~mL\frac{1~g}{1~mL}\frac{1~Kg}{1000~g}~=~0.25~Kg[/tex]
Finalmente para el calculo de la molalidad podemos dividir los dos valores:
[tex]m=\frac{0.42~mol}{0.25~Kg}=1.68[/tex]
Con el valor de la molalidad se puede calcular ΔT al reemplazar los valores:
ΔT=[tex]1.68~\frac{mol}{Kg}*0.52\frac{Kg~^{\circ}C}{mol}=0.87^{\circ}C[/tex]
Si la temperatura de ebullición normal del agua es 100 ºC. Podemos calcular la temperatura final si adicionamos ΔT:
Temperatura final = 100 + 0.87 = 100.87 ºC
Espero sea de ayuda!
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.73 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?
Answer:
The temperature of the air at this given elevation will be 53.32425°C
Explanation:
We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.
Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm
We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T[tex]_2[/tex] ),
P[tex]_1[/tex]V[tex]_1[/tex] / T[tex]_1[/tex] = P[tex]_2[/tex]V[tex]_2[/tex] / T[tex]_2[/tex],
T[tex]_2[/tex] = P[tex]_2[/tex]V[tex]_2[/tex]T[tex]_1[/tex] / P[tex]_1[/tex]V[tex]_1[/tex],
T[tex]_2[/tex] = 0.73 atm [tex]*[/tex] 1.8 L [tex]*[/tex] 298.15 K / 1 atm [tex]*[/tex] 1.2 L = ( 0.73 [tex]*[/tex] 1.8 [tex]*[/tex] 298.15 / 1 [tex]*[/tex] 1.2 ) K = 326.47425 K,
T[tex]_2[/tex] = 326.47425 K = 53.32425 C
A flask contains 6g hydrogen gas and 64 g oxygen at rtp the partial pressure of hydrogen gas in the flask of the total pressure (p)will be
A.2/3p
B.3/5p
C.2/5p
D.1/3p
Answer this with reason
Answer:
B.3/5p
Explanation:
For this question, we have to remember "Dalton's Law of Partial Pressures". This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.
Additionally, we have a proportional relationship between moles and pressure. In other words, more moles indicate more pressure and vice-versa.
[tex]P_i=P_t_o_t_a_l*X_i[/tex]
Where:
[tex]P_i[/tex]=Partial pressure
[tex]P_t_o_t_a_l[/tex]=Total pressure
[tex]X_i[/tex]=mole fraction
With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:
moles of hydrogen gas
The molar mass of hydrogen gas ([tex]H_2[/tex]) is 2 g/mol, so:
[tex]6g~H_2\frac{1~mol~H_2}{2~g~H_2}=~3~mol~H_2[/tex]
moles of oxygen gas
The molar mass of oxygen gas ([tex]O_2[/tex]) is 32 g/mol, so:
[tex]64g~H_2\frac{1~mol~H_2}{32~g~H_2}=~2~mol~O_2[/tex]
Now, total moles are:
Total moles = 2 + 3 = 5
With this value, we can write the partial pressure expression for each gas:
[tex]P_H_2=\frac{3}{5}*P_t_o_t_a_l[/tex]
[tex]P_O_2=\frac{2}{5}*P_t_o_t_a_l[/tex]
So, the answer would be 3/5P.
I hope it helps!
Suppose that 13 mol NO2 and 3 mol H2O combine and react completely. How many moles of the reactant in excess are present after the reaction has completed
Answer:
The number of moles of excess reagent NO₂ that are present after the reaction has completed is 7 moles.
Explanation:
The balanced reaction is:
3 NO₂ + H₂O → 2 HNO₃ + NO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactants and products participate in the reaction:
NO₂: 3 molesH₂O: 1 moleHNO₃: 2 molesNO: 1 molesThe limiting reagent is one that is consumed in its entirety first, determining the amount of product in the reaction. When the limiting reagent ends, the chemical reaction will stop.
In other words, the limiting reagent is that reagent that is consumed first in a chemical reaction, determining the amount of products obtained. The reaction depends on the limiting reagent, because the other reagents will not react when one is consumed.
You can apply the following rule of three: if by stoichiometry of the reaction 3 moles of NO₂ react with 1 mole of H₂O, 13 moles of NO₂ react with how many moles of H₂O?
[tex]moles of H_{2}O=\frac{13 moles of NO_{2}*1 mole of H_{2}O }{3 moles of NO_{2}}[/tex]
moles of H₂O= 4.33 moles
But 4.33 moles of H₂O are not available, 3 moles are available. Since you have less moles than you need to react with 13 moles of NO₂, water H₂O will be the limiting reagent.
To determine the number of moles of excess reagent NO2 that are present after the reaction is complete, you can apply the following rule of three: if by stoichiometry of the reaction 1 moles of H₂O react with 3 mole of NO₂, 3 moles of H₂O react with how many moles of NO₂?
[tex]moles of NO_{2}=\frac{3 moles of NO_{2}*3 mole of H_{2}O }{1 mole of H_{2}O}[/tex]
moles of NO₂= 6 moles
If 6 moles of NO₂ react and 13 moles of the compound are present, the amount that remains in excess is calculated as: 13 moles - 6 moles= 7 moles
The number of moles of excess reagent NO₂ that are present after the reaction has completed is 7 moles.
Given that H2(g)+F2(g)⟶2HF(g)ΔH∘rxn=−546.6 kJ 2H2(g)+O2(g)⟶2H2O(l)ΔH∘rxn=−571.6 kJ calculate the value of ΔH∘rxn for
Answer:
ΔH∘rxn of the reaction is -521.6kJ
Explanation:
Complete question: "Calculate the value of ΔH°rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)"
You can find the ΔH of a reaction by the algebraic sum of similar reactions (Hess's law) as follows:
(1) H₂(g) + F₂(g) ⟶ 2HF(g) ΔH∘rxn=−546.6 kJ
(2) 2H₂(g)+O₂(g)⟶2H₂O(l) ΔH∘rxn=−571.6 kJ
Subtracting 2ₓ(1) - (2)
2ₓ(1) 2H₂(g) + 2F₂(g) ⟶ 4HF(g) ΔH∘rxn=2ₓ−546.6 kJ = -1093.2kJ
-(2) 2H₂O(l) ⟶ 2H₂(g)+O₂(g) ΔH∘rxn=- (-571.6 kJ) = 571.6kJ
2ₓ(1) - (2) 2F₂(g)+ 2H₂O(l) ⟶ 4HF(g) + O₂(g)
H₂(g) are cancelled because are the same in products and reactants
ΔH∘rxn = -1093.2kJ + 571.6kJ
ΔH∘rxn = -521.6
ΔH∘rxn of the reaction is -521.6kJA particular reaction at constant pressure is spontaneous at 390K. The enthalpy change for this reaction is +23.7kJ. What can you conclude about the sign and magnitude of ΔS for this reaction?a. smallb. largec. + smalld. + largee. 0.0
Answer:
+ small
Explanation:
The entropy is obtained from;
∆S= ∆H/T
Where;
∆S= entropy of the system
∆H= enthalpy if the system = +23.7 KJ
T= absolute temperature of the system = 390 K
∆S= 23.7 ×10^3/390 = 60.8 JK^-
There is a small positive change in entropy.
What is the concentration in ppm of 4 g of NaCl dissolved in 100 mL of water?
