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3 years ago

Balance the following chemical equations:

Chemistry

1 answer:

tamaranim1 [39]3 years ago

3 0

Answer:

Explanation

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One

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Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + H2O

Focus on the NO3. This is an odd problem and you usually do not focus on the complex ion. But this one works easiest if you do.

The problem now is going to be the oxygens. There are 2 with the Calcium and only 1 free one going to the water. (The NO3 has been taken care of in the last step).

Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + 2H2O

Count the atoms. I think this equation is balanced.

atom Left Right Result

Ca 1 1 Balanced

O 8 8 Balanced

H 2 + 2 2*2 Balanced

N 2 2 Balanced

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Two

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CH4 + O2====> CO2 + H2O

Start with the hydrogens.

The right side requires a 2

CH4 + O2 ===> CO2 + 2H2O

Now look at the oxygens. There are 4 on the right. and only 2 on the left. You need to multiply O2 by 2

CH4 + 2O2 ===> CO2 + 2H2O

Each side has 1 Carbon 4 hydrogens and 4 oxygens. The equation is balanced.

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Answer: The value of $\Delta H$ for the reaction will be -537 kJ

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

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Molar mass of hydrogen gas = 2 g/mol

Putting values in above equation, we get:

$\text{Moles of hydrogen gas}=\frac{0.647g}{2g/mol}=0.324mol$

We are given:

Energy released for 0.324 moles of hydrogen reacted is 174 kJ

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

By Stoichiometry of the reaction:

When 0.324 moles of hydrogen gas is reacted, the energy evolved is 174 kJ

So, when 1 mole of hydrogen gas will react, the energy evolved will be = $\frac{174kJ}{0.324mol}\times 1mol=537kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

Hence, the value of $\Delta H$ for the reaction will be -537 kJ

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Explanation:

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The ring consist of 3 C atoms, so its name is cyclo propane.

We numbering the atoms of the ring that give the ethyl substituent the low no. (1) and then methyl group take no. (2).

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