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Umnica [9.8K]
2 years ago
14

The reaction below is at equilibrium. What would happen if more sodium were added?

Chemistry
2 answers:
PolarNik [594]2 years ago
8 0

Explanation:

Adding sodium hydroxide (NaOH) will also affect the

position of the equilibrium. While neither sodium ions (Na

+ ) or hydroxide ions (OH - ) are present on either side, the

hydroxide ions will remove H + ions and the equilibrium

will shift to the right hand side to replace the hydrogen

ions that were removed .

if I am right please mark it as brainliest

densk [106]2 years ago
7 0

Answer:

d is your answer please follow me

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The combustion of butane produces heat according to the equation 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn= –5,314 kJ/mol
Dafna11 [192]

Answer:

665 g

Explanation:

Let's consider the following thermochemical equation.

2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol

According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:

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3 years ago
A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution wa
finlep [7]

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

<em>Change in heat:</em>

q = m*S*ΔT

<em>Where q is heat in J,</em>

<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

<em>Molar heat of solution:</em>

3830.87J/0.1048 moles KNO3 =

36554J/mol =

<h3>36.55kJ/mol</h3>

<em />

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