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alexdok [17]
3 years ago
7

Be sure to answer all parts. Calculate ΔH o for the following reaction in two ways, using the data given below. H2(g) + I2(g) →

2HI(g) (a) Using the equation ΔH o rxn = ∑(BE(reactants)) − ∑(BE(products)) kJ/mol (b) Using the equation ΔH o rxn = ∑(nΔH o f (products)) − ∑(mΔH o f (reactants)) kJ/mol Bond BE kJ/mol Substance ΔH o f kJ/mol H―H 436.4 H2(g) 0 I―I 151 I2(g) 61.0 H―I 298.3 HI(g) 25.9
Chemistry
1 answer:
Fantom [35]3 years ago
7 0

Answer:

a) ΔH°rxn = -9.2kJ/mol

b) ΔH°rxn = -9.2kJ/mol

Explanation:

Using Hess's law, you can find ΔH of a reaction from ΔH of formation of the substances involved in the reaction, thus:

ΔH°rxn = ∑(BE(reactants)) − ∑(BE(products))

Or:

ΔH°rxn = ∑(nΔH°f (products)) − ∑(mΔH°f (reactants))

For the reaction:

H₂(g) + I₂(g) → 2HI(g)

a) Using the first equation:

ΔH°rxn = ΔH (H-H) + ΔH (I-I) - 2ΔHBE (H-I)

ΔH°rxn = 436.4kJ + 151kJ - 2×298.3kJ

<em>ΔH°rxn = -9.2kJ/mol</em>

<em />

b) Using the second equation:

ΔH°rxn = 2Δ°f (HI) − ΔH°f (H₂) - ΔH°f (I₂)

ΔH°rxn = 2×25.9kJ - 0kJ - 61.0kJ

<em>ΔH°rxn = -9.2kJ/mol</em>

<em />

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An unbalanced equation is shown. In this reaction, 200.0 g of FeS2 is burned in 100.0 g of oxygen, and 55.00 g of Fe2O3 is produ
Crazy boy [7]
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>

Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.

Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3

To find for the theoretical yield, we first determine the limiting reactant.

100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2

Therefore, the limiting reactant is O2.

Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3

Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
3 0
2 years ago
if i add 25 ml of water to 135 ml of a 0.25 M NaOH solution what will the molarity of the diluted solution be​
DENIUS [597]

Answer:

0.21 M. (2 sig. fig.)

Explanation:

The molarity of a solution is the number of moles of the solute in each liter of the solution. The unit for molarity is M. One M equals to one mole per liter.

How many moles of NaOH in the original solution?

n = c \cdot V,

where

  • n is the number of moles of the solute in the solution.
  • c is the concentration of the solution. c = 0.25 \;\text{M} = 0.25\;\text{mol}\cdot\textbf{L}^{-1} for the initial solution.
  • V is the volume of the solution. For the initial solution, V = 135\;\textbf{mL} = 0.135\;\textbf{L} for the initial solution.

n = c\cdot V = 0.25\;\text{mol}\cdot\textbf{L}^{-1} \times 0.135\;\textbf{L} = 0.03375\;\text{mol}.

What's the concentration of the diluted solution?

\displaystyle c = \frac{n}{V}.

  • n is the number of solute in the solution. Diluting the solution does not influence the value of n. n = 0.03375\;\text{mol} for the diluted solution.
  • Volume of the diluted solution: 25\;\text{mL} + 135\;\text{mL}  = 160\;\textbf{mL} = 0.160\;\textbf{L}.

Concentration of the diluted solution:

\displaystyle c = \frac{n}{V} = \frac{0.03375\;\text{mol}}{0.160\;\textbf{L}} = 0.021\;\text{mol}\cdot\textbf{L}^{-1} = 0.021\;\text{M}.

The least significant number in the question comes with 2 sig. fig. Keep more sig. fig. than that in calculations but round the final result to 2 sig. fig. Hence the result: 0.021 M.

8 0
2 years ago
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Chlorofluorocarbons (CFCs) are no longer used as refrigerants because they destroy the ozone layer. Trichlorofluoromethane (CCl₃
Daniel [21]

Answer:

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

Explanation:

Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.

S=\frac{Q}{T(Kelvins)}

Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol

Temperature at which Trichlorofluoromethan boils , T= 296.95 K

The molar entropy of the evaporation of Trichlorofluoromethan :

=\frac{24.8 kJ/mol}{296.95 K}=0.083516 kJ/mol K = 83.516 J/molK

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

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6. An object is in _____________________________ when its distance from a(n) __________________________is changing. 7. Speed in
strojnjashka [21]

<em>Explanation:</em>

<em>Explanation:An object is in motion when its distance from another object is changing. Whether an object is moving or not depends on your point of view. ... A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point.</em>

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3 years ago
I NEED HELP PLEASE, THANKS! :)
EleoNora [17]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

4 0
2 years ago
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