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stellarik [79]
3 years ago
15

The decomposition of methanol, ch3oh(g), to form ch4(g) and o2(g) absorbs 252.8 kj of heat per mole of oxygen formed. write a ba

lanced thermochemical equation for this reaction.
Chemistry
1 answer:
Svetllana [295]3 years ago
4 0
The thermochemical equation is the chemical equation including the net change of enthalpy (heat).

The chemical equation for the decomposition of methanol to form methane and oxygen is:

2CH3OH --> 2CH4 + O2

The thermochemical equation is:

2CH3OH ---> 2CH4 + O2  - 252.8 kJ

Note that the heat is placed as negative at the right side because it is absorbed during the decomposition, so the environment will have 252.8 kJ less per each mole of O2 produced.

You can equivalently write:

2CH3OH + 252.8 kJ --> 2CH4 + O2
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How many moles of hydrogen gas will be produced if 5.00 moles of zinc reactions with an excess amount of sulfuric acid
harkovskaia [24]

Answer:

Moles of Hydrogen produced is 5 moles

Explanation:

The balanced Chemical equation for reaction between zinc and sulfuric acid is :

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

This equation tells that ; when  1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.

Since sulfuric acid is in excess so Zinc is the limiting reagent

(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)

Limiting reagent (Zn) will decide the amount of Hydrogen produced

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

1\ mole\ zinc\rightarrow 1\ mole\ H_{2}

So,

5\ mole\ zinc\rightarrow 5\ mole\ H_{2}

Hence moles of Hydrogen produced is 5 moles

3 0
3 years ago
For a school event 1/6 of the athletic field is reversed for the fifth -grade classes the reserved part of the field is divided
SVETLANKA909090 [29]

Answer:

\frac{1}{24}

Explanation:

Given:

For a school event, 1/6 of the athletic field is reserved for the fifth -grade classes and the reserved part of the field is divided equally among the 4 fifth grade classes in the school.

To find: fraction of the whole athletic field reserved for each fifth class

Solution:

Fraction of the whole athletic field reserved for four fifth classes = \frac{1}{6}

So, fraction of the whole athletic field reserved for each fifth class = \frac{1}{4}(\frac{1}{6})=\frac{1}{24}

3 0
2 years ago
In the compound silver oxide (Ag₂O), for each one oxygen ion there are two silver ions. If oxygen forms ions with a -2 charge, w
motikmotik

Explanation:

In the compound silver oxide (Ag₂O), for each one oxygen ion there are two silver ions. If ... - did not match any documents.

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3 0
2 years ago
For the reaction of hydrogen with iodine
fenix001 [56]

Answer:

r_{H_2} = \frac{-1}{2} r_{HI}

Explanation:

Hello!

In this case, considering the given chemical reaction:

H_2(g) + I_2(g) \rightarrow 2HI(g)

Thus, by applying the law of rate proportions, we can write:

\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

r_{H_2} = \frac{-1}{2} r_{HI}

Best regards!

8 0
2 years ago
Please Help! Thank You
Ivanshal [37]

Explanation:

the answer is true because I had this question and got it right

3 0
2 years ago
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