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Mazyrski [523]
3 years ago
5

the bunsen burners in your labs are fueled by natural gas which is mostly methane. the thermochemical equation for the combustio

n of methane is
Chemistry
1 answer:
vampirchik [111]3 years ago
5 0

Answer:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Explanation:

The thermochemical equation for the combustion of methane can be obtained as follow:

Methane (CH₄) under goes combustion in the presence of air (O₂) to produce carbon dioxide (CO₂) and water (H₂O). This can be represented in the equation below:

CH₄ + O₂ —> CO₂ + H₂O

Thus, we can balance the equation as follow:

CH₄ + O₂ —> CO₂ + H₂O

There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H₂O as shown below:

CH₄ + O₂ —> CO₂ + 2H₂O

There are a total of 4 atoms of O on the right side and 1 atom on the left side. It can be balance by putting 2 in front of O₂ as shown below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Now the equation is balanced.

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Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t
MariettaO [177]

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

7 0
3 years ago
Calculate the molarity of 1.60L of a solution containing 1.55g of dissolved KBr
Olegator [25]

Answer: the molarity is .00814

Explanation:the formula for molarity is moles over liters but we have grams so we do given over one so 1.55 over 1 multiplied by 1/119.0023 which is tht two atomc masses addes together then you get the moles and divde them by the liters to get the molarity hope this helps god bless

7 0
3 years ago
Which is a nonmetal ?<br><br> 1. Sodium (Na)<br> 2.lithium(Li)<br> 3. Calcium (ca) <br> 4. Carbon(c)
Savatey [412]

Answer: Calcium (ca)

Explanation:

3 0
4 years ago
Read 2 more answers
A chemistry student must write down in her lab notebook the concentration of a solution of potassium chloride. The concentration
DerKrebs [107]

Answer:

3.65 g / ml correct to 3 sig. fig.

Explanation:

The computation of the concentration required is shown below:

As we know that

[A] = mass of solute ÷ volume of solution

Before that first find the mass of solute

Given that

Initial weight = 5.55g

And,

Final weight = 92.7 g

So,

Mass of KCl is

= 92.7 - 5.55

= 87.15 g ~ 87.2 g

Now the KCi is fully dissolved, so the volume is 23.9 ml

So,  concentration is

= 87.2 g ÷ 23.9 ml

= 3.65 g / ml correct to 3 sig. fig.

6 0
3 years ago
Lead oxide is formed when a lead cation that has a charge of 2+ combines with an oxygen anion that has a charge of 2-. Using the
adell [148]
PbO
Not sure how to show crisscross method
Make sure adding the charges together=0 and multiply the elements when necessary to balance charge
4 0
3 years ago
Read 2 more answers
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