Answer:
d
Explanation:
since it is much convenient since the email will not get lost and it's contents will not be forgotten
If H2O acts as an acid in a reaction, what would be its conjugate base?
Answer:
since H2O is an acid, by the Arrhenius definition, it would donate a proton. Thus, the conjugate base is OH~
2. How many joules of heat are released when 32g of water cools down from 71%
specific heat of water is 4.184 J/gºC)
How many kilojoules is this?
he says he doesnt know sorry
The density of mercury is 13.6 g/cm3, What is its density in mg/mm3?
Answer:
Density of mercury is 13600 kg
Calculate the molarity of a 17.5% (by mass) aqueous solution of nitric acid. Select one: a. 2.74 m b. 4.33 m c. 0.274 m d. 3.04 m e. The density of the solution is needed to solve the problem.
Answer:
Option e.
Explanation:
Molarity is the concentration that indicates moles of solute in 1 L of solution.
We have another concentration, percent by mass.
Percent by mass indicates mass of solute in 100 g of solution.
Our solute is HNO₃, our solvent is water.
17.5 g of nitric acid is the mass of solute. We can convert them to moles:
17.5 g . 1mol / 63g = 0.278 moles
We do not have volume of solution. We assume the mass is 100 g because the percent by mass but we need density to state the volume.
Density = Mass / Volume
Mass / Density = Volume
Once we have the volume, we need to be sure the units is in L, to determine molarity
M = mol /L
Cho một thực phẩm có độ ẩm tương đối là 81%, hỏi hoạt độ của nước trong thực phẩm đó là
Given question is:
Given a food with a relative humidity of 81%, what is the water activity in that food?
Explanation:
Water activity in food can be determined by using the below-shown formula:
[tex]water activity=\frac{equilibrium relative humidity}{100}[/tex]
Equilibrium is established between the vapor pressure of food and the surrounding air media.
Thus, for the given food the relative humidity is 81%.
hence, its water activity is
[tex]81/100\\=0.81[/tex]
Gaseous BF3 and BCl3 are mixed in equal molar amounts. All B-F bonds have about the same bond enthalpy, as do all B-Cl bonds. Compare the numbers of microstates to explain why the mixture tends to react to form BF2Cl(g) and BCl2F(g
Solution :
[tex]$BF_3 (g) + BCl_3 (g) \rightarrow BF_2 Cl + BCl_F(g)$[/tex]
Explanation 1 :
Spontaneity of the reaction is based on two factors :
-- the tendency to acquire a state of minimum energy
-- the energy of a system to acquire a maximum randomness.
Now, since there isn't much difference in the bond enthalpies of B-F and B-Cl. So, we can say the major driving factor is tendency to acquire a state of maximum randomness.
Explanation 2 :
A system containing the [tex]\text{"chemically mixed"}[/tex] B halides has a [tex]\text{greater entropy}[/tex] than a system of [tex]$BCl_3$[/tex] and [tex]BF_3[/tex].
It has the same number of [tex]\text{gas phase molecules}[/tex], but more distinguishable kinds of [tex]\text{molecules}[/tex], hence, more microstates and higher entropy.
Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH)2(s) in a solution buffered at a pH of 11.45\
Answer:
Explanation:
a) AgCl(s) in 0.025 M NaCl
Equation: AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)
Initial conc : S O O
equili conc : O S S
NaCl(s) ⇒ Na⁺ (aq) + Cl⁻ (aq)
Initial conc : 0.025 0 0
equili conc : 0 0.025 0.025
Therefore the concentration: Ag⁺ = 6.4 * 10^-9 M, Cl⁻ = 0.025 M
attached below is the detailed solution of the
Which daughter element is produced from the alpha decay of 213 over 85 At ?
A. 213 over 86 Rn
B.217 over 87 Fr
C. 213 over 84 Po
D. 209 over 83 Bi
Answer:
209
83 Bi
Explanation:
213 213 - 4 4
85 At = 85 - 2 Y + 2 He
once formed, how are coordinate covalent bonds different from other covalent bonds?
Answer:
[tex]\boxed {\boxed {\sf {One \ atom \ donates \ both \ electrons \ in \ a \ pair}}}[/tex]
Explanation:
A covalent bond involves the sharing of electrons to make the atoms more stable, and so they satisfy the Octet Rule (8 valence electrons).
Typically each atom contributes an electron to form an electron pair. This is a single bond. There are also double bonds (two pairs of electrons), triple bonds (three pairs of electrons), and coordinate covalent bonds.
Sometimes, to satisfy the Octet Rule and achieve stability, one atom contributes both of the electrons in an electron pair. This is different from other covalent bonds because usually each of the 2 atoms contributes an electron to make a pair.
complete the following steps.
Remember to follow lower numbered rules first.
Na2CO3(aq) + Pb(OH)2(aq) → NaOH (?) + PbCO3(?)
a. Write a balanced chemical equation. (1 pt)
b. If a reaction occurs, write the balanced
chemical equation with the proper states of matter
(i.e. solid, liquid, aqueous) filled in. If no reaction
occurs, write “No reaction.” (1 pt)
c. If a reaction occurs, write the net ionic equation
for the reaction. If no reaction occurs, write "no
reaction.” (1 pt)
Answer:
See explanation
Explanation:
a) The balanced reaction equation is;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH + PbCO3
b) When we include states of matter;
Na2CO3(aq) + Pb(OH)2(aq) -----> 2 NaOH(aq) + PbCO3 (s)
c) Complete ionic equation;
2Na^+(aq) + CO3^2-(aq) + Pb^2+(aq) + 2OH^-(aq) ----> 2Na^+(aq) + 2OH^-(aq) + PbCO3(s)
Net Ionic equation;
Pb^2+(aq) + CO3^2-(aq) ----> PbCO3(s)
Use dimensional analysis to solve the following problems. Pay attention to correct use of units and correct use of significant figures in calculations. Please show work!
Convert 3.00 x 10^21 atoms of copper to moles.
Convert 2.25 x 10^18 molecules of carbon dioxide to moles.
Answer:
1) 0.00498 mol Cu.
2) 0.00000374 mol CO₂
Explanation:
Question 1)
We want to convert 3.00 * 10²¹ copper atoms into moles. Note that 3.00 is three significant figures.
Recall that by definition, one mole of a substance has exactly 6.022 * 10²³ amount of that substance. In other words, we have the ratio:
[tex]\displaystyle \frac{1\text{ mol}}{6.022\times 10^{23} \text{ Cu}}[/tex]
We are given 3.00 * 10²¹ Cu. To cancel out the Cu, we can multiply it by our above ratio with Cu in the denominator. Hence:
[tex]\displaystyle 3.00 \times 10^{21} \text{ Cu} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} \text{ Cu}}[/tex]
Cancel like terms:
[tex]=\displaystyle 3\times 10^{21} \cdot \frac{1\text{ mol Cu}}{6.022\times 10^{23} }[/tex]
Simplify:
[tex]\displaystyle = \frac{3\text{ mol Cu}}{6.022 \times 10^{2}}[/tex]
Use a calculator:
[tex]= 0.004981... \text{ mol Cu}[/tex]
Since the resulting answer must have three significant figures:
[tex]= 0.00498\text{ mol Cu}[/tex]
So, 3.00 * 10²¹ copper atoms is equivalent to approximately 0.00498 moles of copper.
Question 2)
We want to convert 2.25 * 10¹⁸ molecules of carbon dioxide into moles. Note that 2.25 is three significant digits.
By definition, there will be 6.022 * 10²³ carbon dioxide molecules in one mole of carbon dioxide. Hence:
[tex]\displaystyle \frac{6.022 \times 10^{23} \text{ CO$_2$}}{1\text{ mol CO$_2$}}[/tex]
To cancel the carbon dioxide from 2.25 * 10¹⁸, we can multiply it by the above ratio with the carbon dioxide in the denominator. Hence:
[tex]\displaystyle 2.25\times 10^{18} \text{ CO$_2$} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23} \text{ CO$_2$}}[/tex]
Cancel like terms:
[tex]\displaystyle= 2.25\times 10^{18} \cdot \frac{1\text{ mol CO$_2$}}{6.022\times 10^{23}}[/tex]
Simplify:
[tex]\displaystyle = \frac{2.25 \text{ mol CO$_2$}}{6.022\times 10^5}}[/tex]
Use a calculator:
[tex]=0.000003736...\text{ mol CO$_2$}[/tex]
Since the resulting answer must have three significant figures:
[tex]= 0.00000374\text{ mol CO$_2$}[/tex]
So, 2.25 * 10¹⁸ molecules of carbon dioxide is equivalent to approximately 0.00000374 moles of carbon dioxide.
Answer:
Explanation:
by definition, 1 mole contains 6.02 x 10^23 of atoms (for elements) or molecules (for compounds)
3.00 x 10^21 atoms of copper / 6.02 x 10^23 of atoms
= 0.004983 moles of copper
= 4.98 x 10^(-3) moles of copper
2.25 x 10^18 molecules of carbon dioxide / 6.02 x 10^23 of molecules
= 0.000003737 moles of carbon dioxide
= 3.74 x 10^(-6) moles of carbon dioxide
What force is behind us when we ride a bike?
Answer:
gravity, ground, friction, rolling resistance, and air resistance.
Why are prefixes not needed in naming ionic compounds?
Answer:
when naming ionic compounds — those are only used in naming covalent molecular compounds. Do NOT use prefixes to indicate how many of each element is present; this information is implied in the name of the compound. since iron can form more than one charge. Ionic Compounds Containing a Metal and a Polyatomic Ion.
10.0 g of Mg are reacted with 95.0 g of I2. If 27.5 g of magnesium iodide are obtained, what is the percent yield
Answer:
% yield of reaction is 26.4
Explanation:
The reaction is:
Mg + I₂ → MgI₂
Our reactants are magnessium and iodine. We determine the moles of each to find the limiting reactant:
10 g . 1mol / 24.3 g = 0.411 moles of Mg
95 g . 1mol / 253.8g = 0.374 moles of I₂
Ratio is 1:1. For 1 mol of Mg we need 1 mol of iodine
For 0.411 moles, we need the same amount, but we only have 0.374 moles of iodine, that's why the gas is the limiting reactant.
As ratio is 1:1 again, 0.374 moles of iodine can produce 0.374 moles of MgI₂
We determine the mas (theoretical yield): 0.374 mol . 278.1 g/mol = 104 g
To calculate the percent yield:
% yield = (yield produced /theoretical yield) . 100
% yield = (27.5 g/ 104g) . 100 = 26.4
When we test sucrose with seliwanoff's test what would the result be positive or negative ? Before and after hydrolysis of sucrose .
I need more explain?
In an experiment 25.0 mL of 0.100 M KI was diluted to 50.0 mL. Calculate the molarity of the diluted solution
Answer:
The molarity is "0.050 M".
Explanation:
The given values are:
M1 = 0.100 M
M2 = ?
V1 = 25.0 mL
V2 = 50.0 mL
As we know,
⇒ [tex]M1\times V1=M2\times V2[/tex]
Or,
⇒ [tex]M2=\frac{M1\times V1}{V2}[/tex]
By putting the values, we get
[tex]=\frac{0.100\times 25}{50}[/tex]
[tex]=\frac{2.5}{50}[/tex]
[tex]=0.05 \ M[/tex]
A hypothetical A-B alloy of composition 53 wt% B-47 wt% A at some temperature is found to consist of mass fractions of 0.5 for both and phases. If the composition of the phase is 92 wt% B-8 wt% A, what is the composition of the phase
Answer:
the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Explanation:
Given the data in the question;
Co = 53 or [ 53 wt% B-47 wt% A ]
W∝ = 0.5 = Wβ
Cβ = 92 or [ 92 wt% B-8 wt% A ]
Now, lets set up the Lever rule for W∝ as follows;
W∝ = [ Cβ - Co ] / [ Cβ - C∝ ]
so we substitute our given values into the expression;
0.5 = [ 92 - 53 ] / [ 92 - C∝ ]
0.5 = 39 / [ 92 - C∝ ]
0.5[ 92 - C∝ ] = 39
46 - 0.5C∝ = 39
0.5C∝ = 46 - 39
0.5C∝ = 7
C∝ = 7 / 0.5
C∝ = 14 or [ 14 wt% B-86 wt% A ]
Therefore, the composition of the ∝ phase C∝ = 14 or [ 14 wt% B-86 wt% A ]
Classify each molecule as an alcohol, ketone, or aldehyde based on its name. Propanone (acetone) Choose... Ethanal Choose... 3-phenyl-2-propenal Choose... Butanone Choose... Ethanol Choose... 2-propanol Choose...
Answer:
1.) Propanone (ketone)
2.) Ethanal( aldehyde)
3.) 3-phenyl-2-propenal (aldehyde)
4.) Butanone (ketone)
5.) Ethanol ( alcohol)
6.) 2-propanol (alcohol)
Explanation:
In organic chemistry, ALCOHOL ( also known as alkanol) are compounds in which hydroxyl groups are linked to alkyl groups. They can be considered as being derived from the corresponding alkanes by replacing the hydrogen atoms with hydroxyl groups. The hydroxyl group is the functional group of the alcohol as it is responsible for their characteristic chemical properties. A typical example of alcohol is ethanol and 2-propanol.
Alkanals or ALDEHYDES have the general formula RCHO while alkanones or KETONES have the general formula RR'CO where R and R' may be alkyl or aryl groups. The main similarity between these two classes of compounds is the presence of the carbonyl group. In aldehydes, there is a hydrogen atom attached to the carbon In the carbonyl group while there is none on the ketones.
Some common examples of ketones are Propanone, Butanone while examples of aldehydes are Ethanal and 3-phenyl-2-propenal
According to the kinetic theory, all matter is made of moving particles, which measurement of matter is directly proportional to the
average kinetic energy of the particles?
When 250. mg of eugenol, the molecular compound responsible for the odor of oil of cloves, was added to 100. g of camphor, it lowered the freezing point of camphor by 0.62 8C. Calculate the molar mass of eugenol.
Answer:
Molar mass for eugenol is 161.3 g/mol
Explanation:
This question talks about freezing point depression:
Our solute is eugenol.
Our solvent is camphor.
Formula to state the freezing point depression difference is:
ΔT = Kf . m . i where
ΔT = Freezing T° of pure solvent - Freezing T° of solution
In this case ΔT = 0.62°C
Kf for camphor is: 37°C /m
As eugenol is an organic compund, i = 1. No ions are formed.
To state the molar mass, we need m (molal)
Molal are the moles of solute in 1kg of solvent. Let's replace data:
0.62°C = 40 °C/m . m . 1
0.62°C / 40 m/°C = 0.0155 m
We convert mass of camphor from g to kg = 100 g . 1kg / 1000g = 0.1 kg
0.0155 molal = moles of solute / 0.1 kg
0.0155 m/kg . 0.1 kg = 0.00155 moles
We know that these moles are contained in 250 mg, so the molar mass will be:
0.25 g / 0.00155 mol = 161.3 g/mol
Notice, we convert mg to g, for the units!
Which intermolecular force plays a pivotal role in biological molecules such as proteins and DNA ?
•hydrogen bonding
•dispersion force
•dipole-dipole force
•Ion-dipole force
Answer:
hydrogen bonding
Explanation:
just took the test :D
Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.
In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?
For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)
Answer:
Mass of C = 47.37g
Mass of H = 10.59g
Mass of O = 42.04g
The total mass of these elements is 100g, taking a proportion of their molar masses.
C = 47.37/12= 3.95
H = 10.59/1 = 10.59
O = 42.04/16= 2.63.
Dividing through with the smallest proportion which is 2.63
C=3.95/2.63 = 1.5
H =10.59/2.63 =4
O = 2.63/2.63= 1
Multiplying through by 2 to get a whole number.
C = 1.5x2 = 3
H= 4x2 = 8
O = 1x2= 2
The empirical formula is C3H6O2
(Empirical formula)n= molecular mass
(C3H8O2)n =228.276
(12x3 +8+16x2)n= 228.276
76n = 228.276
n = 228.276/76
n = 3
Molecular formula = Empirical formula
=(C3H8O2)3 = C9H24O6
The molecular formula is C9H24O6
(c) m X is an ion in which group of the periodic table is the element from which X is formed?
Explanation:
Iron has 2 atoms and 3atoms.
So,X=2,3
How many atoms are present in 0.45 moles of P4010
Answer:
80g
Explanation:
mass oxygen present in 1 mole of p4010
16×10=160gm
similarly
for 0.5 moles of p4010 160/2= 80gm
The number of atoms present in 0.45 moles of P₄O₁₀ is 1.08 x 10²³ atoms.
To determine the number of atoms, we use Avogadro's number, which states that there are approximately 6.022 x 10²³ particles (atoms, molecules, or formula units) in one mole of a substance.
In this case, we are given 0.45 moles of P₄O₁₀. To calculate the number of atoms, we multiply the number of moles by Avogadro's number:
Number of atoms = 0.45 moles P₄O₁₀ x (6.022 x 10²³ atoms / 1 mole)
Number of atoms = 2.7139 x 10²³ atoms
Rounding to three significant figures, the number of atoms present in 0.45 moles of P₄O₁₀ is approximately 1.08 x 10²³ atoms.
To learn more about atoms here
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Using a balanced chemical equation, and 2.50 g of sodium hydrogen carbonate as the reactant,
what is the expected (theoretical) yield of sodium carbonate (grams)? The Formula Weight (FW) of
sodium hydrogen carbonate is 84.01 g and sodium carbonate is 105.99 g.
Answer:
1.58 g
Explanation:
Step 1: Write the balanced equation
2 NaHCO₃ ⇒ Na₂CO₃ + H₂O + CO₂
Step 2: Calculate the moles corresponding to 2.50 g of NaHCO₃
The molar mass of NaHCO₃ is 84.01 g/mol.
2.50 g × 1 mol/84.01 g = 0.0298 mol
Step 3: Calculate the moles of Na₂CO₃ produced
The molar ratio of NaHCO₃ to Na₂CO₃ is 2:1. The moles of Na₂CO₃ produced are 1/2 × 0.0298 mol = 0.0149 mol
Step 4: Calculate the mass corresponding to 0.0149 moles of Na₂CO₃
The molar mass of Na₂CO₃ is 105.99 g/mol.
0.0149 mol × 105.99 g/mol = 1.58 g
In the given range,at what temperature does oxy gen have the highest solubility?
8moles of Na2Cr2O2 is how much mass
[tex] \boxed{\boxed{\mathfrak{ 1\: mole \:of \:Na_2Cr_2O_2\: = \:it's \:Gram\: Mol. \: mass}} }[/tex]
[tex]\underline{ \mathfrak{ Gram \:molecular \:mass \:of \: \red{ Na_2Cr_2O_2}}}[/tex]
= 2 × 23 + 2 × 52 + 2 × 16
= 182 grams
1 mole of [tex]Na_2Cr_2O_2[/tex] weighs = 182 g
8 moles weigh = 8× 182
=[tex] \mathfrak{\blue {\boxed{\underline {1456 \: grams}}}} [/tex]
or
[tex] \mathfrak{\blue {\boxed{\underline {1. 46 \:kg }}}} [/tex]
Solid potassium chlorate (KClO3)(KClO3) decomposes into potassium chloride and oxygen gas when heated. How many moles of oxygen form when 48.1 gg completely decomposes
Answer:
0.59 mol O₂
Explanation:
The balanced chemical equation for the decomposition of potassium chlorate (KClO₃) to produce potassium chloride (KCl) and oxygen gas (O₂) is the following:
2 KClO₃ → 2 KCl + 3 O₂
According to the equation, 3 moles of O₂ are produced from 2 moles of KClO ⇒ conversion factor: 3 mol O₂/2 mol KClO₃
Now, we calculate the number of moles of KClO₃ there is in 48.1 g, by dividing the mass into the molecular weight (Mw) of O₂:
Mw(KClO₃) = 39.1 g/mol + 35.4 g/mol + (16 g/mol x 3) = 122.5 g/mol
moles KClO₃ = mass KClO₃/Mw(KClO₃) = 48.1 g/(122.5 g/mol) = 0.3926 mol KClO₃
Finally, we multiply the moles of KClO₃ by the conversion factor to calculate the moles of O₂ produced:
0.3926 mol KClO₃ x 3 mol O₂/2 mol KClO₃ = 0.59 mol O₂
Consider the reaction C4H10O + NaBr + H2SO4 → C4H9Br + NaHSO4 + H2O. If 45.0 g of C4H10O reacts with 67.1 g of NaBr and 97.0 g of H2SO4to yield 60.0 g of C4H9Br, calculate the percent yield of the reaction.
Answer:
Percent yield = 72.07 %
Explanation:
Our reaction is:
C₄H₁₀O + NaBr + H₂SO₄ → C₄H₉Br + NaHSO₄ + H₂O
It is correctly balanced.
Let's determine which is the limiting reagent:
45 g . 1 mol / 74 g = 0.608 moles of C₄H₁₀O
67.1 g . 1 mol / 102.9 g = 0.652 moles of NaBr
97 g . 1 mol / 98 g = 0.990 moles of sulfuric acid
Ratio is always 1:1, so for 1 mol of NaBr and 1 mol of sulfuric acid we need 1 mol of C₄H₁₀O. We have 0.652 moles of NaBr, we need the same amount of C₄H₁₀O and we have 0.990 moles of acid, we need the same amount of C₄H₁₀O; we only have 0.608 moles, that's why C₄H₁₀O is the limiting reactant, there's no enough C₄H₁₀O.
Ratio is also 1:1, between reactant and product.
1 mol of C₄H₁₀O produces 1 mol of C₄H₉Br
Then, 0.608 moles will produce 0.608 moles of C₄H₉Br
We convert moles to mass: 0.608 mol . 136.9 g/mol = 83.25 g
That's the 100 % yield reaction
Percent yield = (Yield produced / Theoretical yield) . 100
Percent yield = (60 g / 83.25 g) . 100 = 72.07 %
79.1,3-Butadiene molecule contains how many sigma and pi bond
3 sigma and 3 pieee
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