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3 years ago

Must show work for questions. Please also explain, I don't understand.

Chemistry

1 answer:

prisoha [69]3 years ago

3 0

Answer:

11. 227 kJ·mol⁻¹; 12. 587 kJ

Step-by-step explanation:

11. Calculate ΔH°f

The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is

ΔH°r = ΣmΔH°f(products) - ΣnΔH°f(reactants)

2C₂H₂ + 5O₂ ⟶ 4CO₂ + 2H₂O; ΔH°c = -1299 kJ·mol⁻¹

ΔH°f/kJ·mol⁻¹: 2x 0 -393 -286

2(-1299) = 4(-393) + 2(-286) - 2x

-2598 = -1572 - 572 - 2x

-2598 = -2144 -2x

-454 = -2x

x = 227 kJ·mol⁻¹

ΔH°f for acetylene is 227 kJ·mol⁻¹

12. Hess's Law

Warning: It is almost impossible to solve a problem like this by trial and error. You must have a strategy beforehand.

We have five equations:

(I) Ca + 2C ⟶ CaC₂; ΔH = -62.8 kJ

(II) Ca + ½O₂ ⟶ CaO; ΔH = -635.5 kJ

(III) CaO + H₂O ⟶ Ca(OH)₂; ΔH = -653.1 kJ

(IV) C₂H₂ + ⁵/₂O₂ ⟶ 2CO₂ + H₂O; ΔH = -1300 kJ

(V) C + O₂ ⟶ CO₂; ΔH = -393.51 kJ

From these, we must devise the target equation:

(VI) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂; ΔH = ?

Here comes the strategy. Remember, you can use each of the given equations only once.

Start with the target equation. It has CaC₂ on the left, so you need an equation containing CaC₂ on the left.

Reverse Equation (I). When you reverse an equation, you change the sign of its ΔH.

(VII) CaC₂ ⟶ Ca + 2C; ΔH = 62.8 kJ

Equation (VII) has Ca on the right, and that is not in the target. You need an equation with Ca on the left to cancel it. Add Equation (II)

(VIII) Ca + O₂ ⟶ CaO; ΔH = -1271.0 kJ

Equation (VII) also has 2C on the right, and there is no C in the target. We need an equation with 2C on the left to cancel it.

Double Equation (V). When you double an equation, you double its ΔH.

(IX) 2C + 2O₂ ⟶ 2CO₂; ΔH = -587.02 kJ

Now, we must eliminate the CaO from Equation (VII). Add Equation (III).

(X) CaO + H₂O ⟶ Ca(OH)₂; ΔH = -653.1 kJ

Equation (VIII) has 2CO₂ on the right. You need an equation with 2CO₂ on the left.

Reverse Equation (IV).

(XI) 2CO₂ +H₂O ⟶ C₂H₂ + ⁵/₂O₂ ; ΔH = 2600 kJ

Now, we add all five equations, cancelling substances that appear on both sides of the reaction arrow. All the formulas in italics have been cancelled

(VII) CaC₂ ⟶ Ca + 2C; ΔH = 62.8 kJ

(VIII) Ca + ½O₂ ⟶ CaO; ΔH = -635.5 kJ

(IX) 2C + 2O₂ ⟶ 2CO₂; ΔH = -787.02 kJ

(X) CaO + H₂O ⟶ Ca(OH)₂; ΔH = -653.1 kJ

(XI) 2CO₂ + H₂O ⟶ C₂H₂ + ⁵/₂O₂ ; ΔH = 2600 kJ

(VI) CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂; ΔH = 587 kJ

ΔH = 587 kJ

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ASAP , 8.81 g Carbon

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The empirical formula : C₂Cl₇

The molecular formula : C₁₀Cl₃₅

<h3>Further explanation</h3>

Given

8.81 g Carbon

91.2 g Chlorine

Molar Mass: 1362.5 g/mol

Required

The empirical formula and molecular formula

Solution

Mol ratio :

C = 8.81 g : 12.011 g/mol =0.733

Cl = 91.2 g : 35,453 g/mol = 2..572

Divide by 0.733

C : Cl = 1 : 3.5 = 2 : 7

The empirical formula : C₂Cl₇

(The empirical formula)n = the molecular formula

(C₂Cl₇)n = 1362.5

(2x12.011+7x35.453)n=1362.5

(272.193)n=1362.5

n = 5

6 0

3 years ago

Precipitation reactions always occur when two aqueous solutions are mixed (T/F)

Studentka2010 [4]

Answer: The given statement is false.

Explanation:

Precipitation reaction is defined as the chemical reaction in which two aqueous solution upon mixing together results in the formation of an insoluble solid.

For example, $NaCl(aq) + AgNO_{3} \rightarrow NaNO_{3}(aq) + AgCl(s)$

Here AgCl is present in solid state so, it is the precipitate.

But it is not necessarily true that two aqueous solutions will always result in the formation of a precipitate.

For example, $NaCl(aq) + KNO_{3}(aq) \rightarrow KCl(aq) + NaNO_{3}(aq)$

Hence, we can conclude that the statement precipitation reactions always occur when two aqueous solutions are mixed, is false.

7 0

3 years ago

Decreasing the temperature of the reaction 3H2 + N2 2NH3. In this reaction, the product absorbs heat. WHICH WAY WILL THE REACTIO

Alex

Answer:

the reaction will shift towards the “heat”—shifts to the left

Explanation:

To summarize:

o If temperature increases (adding heat), the reaction will shift away from the “heat” term and go in the

endothermic direction.

o If temperature decreases (removing heat), the reaction will shift towards the “heat” term and go in the

exothermic direction.

o NOTE: The endothermic direction is always away from the “heat” term and the exothermic direction is

towards the “heat” term.

Therefore the reaction will shift towards the “heat”—shifts to the left

3 0

4 years ago

Tin (II) fluoride, formerly found in many kinds of toothpaste, is formed in this reaction: Sn (s) + 2HF (g) ——> SnF2 (s) + H2

Readme [11.4K]

1.34 L of HF

Explanation:

We have the following chemical reaction:

Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)

First we calculate the number of moles of SnF₂:

number of moles = mass / molecular weight

number of moles of SnF₂ = 5 / 157 = 0.03 moles

From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.

Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:

number of moles = volume / 22.4 (L/mole)

volume of HF = number of moles × 22.4

volume of HF = 0.06 × 22.4 = 1.34 L

Learn more about:

problems with gases at STP

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3 years ago

Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn

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