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2 years ago

Using the following bond energies:

Chemistry

1 answer:

uranmaximum [27]2 years ago

4 0

Answer:

The answer is "Option b".

Explanation:

Given equation and value:

$\Rightarrow C_2H_2 \ (g)+ \frac{5}{2}O_2 \ (g) \longrightarrow 2CO_2\ (g)+H_2O\ (g)$

$\left\begin{array}{ccc}C \equiv C&839\\C-H &413\\O=O &495\\ C=O &799\\ O-H &467 \end{array}\right \\\\$

calculating equation value:

$\Rightarrow 839 + 2(413) +\frac{5}{2} (495) \longrightarrow 2(2)799+2 \times 467 +E$

$\Rightarrow 839 + 826 +1237.5 = 3196+934 +E\\\\\Rightarrow 2902.5 =4130 +E\\\\\Rightarrow 2902 = 4130 +E\\\\\Rightarrow 2902 - 4130 +E\\\\\Rightarrow E = -1228 \ KJ \\\\$

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worty [1.4K]

C is the answer.
The temperature T<span> in degrees Celsius (°C) is equal to the temperature </span>T<span> in Kelvin (K) minus 273</span>°.

6 0

3 years ago

Read 2 more answers

A 33.0 mL sample of 1.15 M KBr and a 59.0 mL sample of 0.660 M KBr are mixed. The solution is then heated to evaporate water unt

Oksi-84 [34.3K]

Answer:

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

Explanation:

<u>Step 1:</u> Data given

Sample 1: The 1.15 M sample has a volume of 33.O mL

Sample 2: The 0.660 M sample has a volume of 59.0 mL

Molar mass of KBr = 119 g/mol

Molar mass of AgNO3 = 169.87 g/mol

<u>Step 2:</u> Calculate number of moles for both samples

Number of moles = Molarity * Volume

Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles

Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles

Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles

<u>Step 3:</u> Calculate total mass

mass = Number of moles * Molar mass

mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)

<u>Step 4</u>: Calculate moles of AgBr

AgNO3 reacts with KBr

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3

So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3

<u>Step 5:</u> Calculate mass of silver nitrate

mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3

mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams

We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution

8 0

3 years ago

Consider the following reaction:

iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0

3 years ago

Does neon have a larger atomic radius than argon

Lera25 [3.4K]

Answer:

Explanation:

Argon has a larger atomic radius

6 0

2 years ago

Determine the number of hydrogen atoms in the following compound 3Cu(C2H3O2)2

Charra [1.4K]

Answer:5

Explanation:(H3) is 3 (H3)2 is 5

6 0

2 years ago