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2 years ago

Question 14 of 32

Chemistry

1 answer:

ser-zykov [4K]2 years ago

6 0

The equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.

<h3>What is the equilibrium expression?</h3>

The equilibrium expression shows how much of the reactants are converted into products. If the equilibrium constant is large and positive, most of the reactants have been converted into products.

Thus, the equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.

Learn more about equilibrium constant:brainly.com/question/10038290

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What is safer to drink 113 ml of HCl or 244 ml of NaOH

skad [1K]

Answer:

bleach

Explanation:

idrk

5 0

2 years ago

What is the molality of a solution that contains 1.34 moles of NaCl in 2.47 kg of solvent

dsp73

The molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

<u>Explanation:</u>

Molality is the measure of how much of amount of solute is dissolved in the solvent. So it is calculated as the ratio of moles of solute to the grams of solvent.

$\text {Molality}=\frac{\text {Moles of solute}}{\text {Mass of solvent}}$

As in this case, the solute is NaCl and solvent is unknown. So the moles of solute is given as 1.34 moles and the mass of solvent is given as 2.47 kg.

Hence, $\text { molality }=\frac{1.34}{2.47}= 0.54 \mathrm{M}$

Thus, the molality is 0.54 M when 1.34 moles of NaCl is present in 2.47 kg of solvent.

8 0

3 years ago

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

Neopentyl alcohol, (ch3)3cch2oh, cannot be dehydrated to an alkene without rearrangement. what is the chief product of dehydrati

emmainna [20.7K]

Dehydration is removal of water.

In alcohols dehydration is α-β elimination or 1,2 elimination, it means the hydroxyl group will be removed from α-carbon while the hydrogen will be removed from near by carbon.

In case of neopentyl alcohol there is no β hydrogen present on the β carbon [as shown in figure].

The only possible way for it to undergo dehydration is by rearrangement.

The process or mechanism can be understood as:

so the chief product is 2-methylbut-2-ene

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3 years ago

The step by step direction of a experiment are found in...

blondinia [14]

Answer:

I believe it the Data Table

Explanation:

because that the final step of an experiment; recording your data throughout the experiment, and that where you recorded your steps and information throughout the experiment.

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2 years ago