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2 years ago

What is the value of delta Hrxn for this equation:

Chemistry

1 answer:

Ratling [72]2 years ago

6 0

Answer:

The ΔHrxn for the above equation = 179 kJ/mol

Explanation:

The reaction bond enthalpies are for the reactant;

3 × N-H = 3 × 390 = 1,170 kJ/mol

2 × O=O = 2 × 502 = 1004 kJ/mol

The reaction bond enthalpies are for the product;

3 × N-O = 3 × 201 = 603 kJ/mol

3 × O-H = 3 × 464 = 1,392 kJ/mol

The ΔHrxn for the above equation is therefore;

ΔHrxn = 1,170 + 1,004 - (603 + 1,392) = 179 kJ/mol

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Answer:

Law of Lateral Continuity The Grand Canyon.

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he same rock layers on opposite sides of the canyon. The matching rock layers were deposited at the same time, so they are the same age.

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3 years ago

Which represents the balanced nuclear equation for the beta plus decay of C-11?

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Nuclear reaction: ¹¹C → ¹¹B + e⁺(positron) + ve(electron neutrino).
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2 years ago

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A metal oxide with the formula mo contains 15.44% oxygen. in the box below, type the symbol for the element represented by m.

lana66690 [7]

Answer: The element represented by M is Strontium.

Explanation:

Let us consider the molar mass of metal be 'x'.

The molar mass of MO will be = Molar mass of oxygen + Molar mass of metal = (16 + x)g/mol

It is given in the question that 15.44% of oxygen is present in metal oxide. So, the equation becomes:

$\frac{15.44}{100}\times (x+16)=16g/mol\\\\(x+16)=\frac{16g/mol\times 100}{15.44}\\\\x=(103.626-16)g/mol\\\\x=87.62g/mol$

The metal atom having molar mass as 87.62/mol is Strontium.

Hence, the element represented by M is Strontium.

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3 years ago

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A chemist conducted a study on ammonia NH3(g) into hydrogen gas and nitrogen gas. This reaction is represented by the following

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Answer:

Explanation:

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2 years ago

How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me

Georgia [21]

First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.

Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min

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3 years ago