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Tamiku [17]
3 years ago
9

A person uses 770 kcal on a long hike. Calculate the energy used for the hike in each of the following energy units.

Chemistry
1 answer:
Pavlova-9 [17]3 years ago
5 0

Explanation:

1 Cal of energy is equivalent to 4.184 Joules.

A.

770Cal of energy = 3221.680 joules

Therefore, 780kCal of energy = 3221680 Joules.

B.

3200000 Joules (in 2 s.f).

C.

I000 joules = 1 kJoule

3221680 joules = 3221.689 kJoules.

D.

3200 kJ (in 2 s.f).

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Answer:

28.9 g

Explanation:

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.  

<em>Gather all the information in one place</em> with molar masses above the formulas and masses below them.  

M_{r}:     159.69    28.01

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Mass/g:  55.0

1. Use the molar mass of Fe₂O₃ to calculate the moles of Fe₂O₃.

\text{Moles of Fe$_{2}$O$_{3}$} =\text{55.0 g Fe$_{2}$O$_{3}$} \times \frac{\text{1 mol Fe$_{2}$O$_{3}$}}{\text{159.69 g Fe$_{2}$O$_{3}$}}= \text{0.3444 mol Fe$_{2}$O$_{3}$}

2. Use the molar ratio of CO:Fe₂O₃ to calculate the moles of CO.

\text{Moles of CO} = \text{0.3444 mol Fe$_{2}$O$_{3}$} \times \frac{\text{3 mol CO}}{\text{1 mol Fe$_{2}$O$_{3}$}}= \text{1.033 mol CO}

3.Use the molar mass of CO to calculate the mass of CO.

\text{Mass of CO} = \text{1.033 mol CO}  \times \frac{\text{28.01 g CO} }{\text{1 mol CO}}= \textbf{28.9 g CO}  

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Acetylene gas (ethyne; HC = CH) burns in an oxyacetylene torch to produce carbon dioxide and water vapor. The heat of reaction f
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The mass of CO2 produced by 26g of acetylene is 88g.

Given ,

In an oxyacetylene torch, acetylene gas (ethyne; HCCH) burns to produce carbon dioxide and water vapour.

The acetylene combustion reaction is given by,

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Heat of reaction for acetylene combustion = 1259kj/mol

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2 moles of CO2 have a molecular mass of 88g.

On combustion, 1 mole of acetylene yields 2 moles of CO2.

Thus, 26g of acetylene produces 88g of CO2 when burned.

As a result, the mass of carbon dioxide produced by 26g of acetylene is 88g.

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