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3 years ago

Consider the balanced chemical equation for the combustion of methane (CH4).

Chemistry

2 answers:

Mrac [35]3 years ago

5 0

<h3>Answer:</h3>

89.6 L of O₂

<h3>Solution:</h3>

The balanced chemical equation is as,

CH₄ + 2 O₂ → CO₂ + 2 H₂O

As at STP, one mole of any gas (Ideal gas) occupies exactly 22.4 L of Volume. Therefore, According to equation,

44 g ( 1 mol) CO₂ is produced by = 44.8 L (2 mol) of O₂

So,

88 g CO₂ will be produced by = X L of O₂

Solving for X,

X = (88 g × 44.8 L) ÷ 44 g

X = 89.6 L of O₂

KIM [24]3 years ago

3 0

Took a guess and got it right!

D) 89.6 L

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4 years ago

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3 years ago

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

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Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

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F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

<u><em>2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.</em></u>

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$

mass of CaF₂= 1000 g

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

$moles=\frac{1000 grams*1 mole}{78 grams}$

moles=12.82

<u><em>12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution</em></u>

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3 years ago