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Lelu [443]
3 years ago
12

SOLVE The density of a gas is 0.68 g/mL. What amount of space would 23.8 g of this gas occupy?

Chemistry
1 answer:
Contact [7]3 years ago
6 0

Answer:

<h2>The answer is 35 mL</h2>

Explanation:

Density of a substance can be found by using the formula

Density(\rho) =  \frac{mass}{volume}

From the question we are finding the amount of space the gas will occupy that's the volume of the gas

Making volume the subject we have

volume =  \frac{mass}{Density}

From the question

mass = 23.8 g/mL

Density = 0.68 g/mL

Substitute the values into the above formula and solve

That's

volume =  \frac{23.8}{0.68}

We have the final answer as

<h3>35 mL</h3>

Hope this helps you

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3 years ago
A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp
MArishka [77]

Answer:

V = 11.2L are produced

Explanation:

... <em>Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..</em>

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

<em>Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.</em>

<em />

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

<em>Moles Al-Molar mass: 26.98g/mol-:</em>

27g * (1mol / 26.98g) = 1mol of Al

<em>Moles HCl:</em>

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and <em>HCl is limiting reactant.</em>

<em />

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

<em>Where V is volume</em>

<em>n are moles: 0.5mol</em>

<em>R is gas constant: 0.082atmL/molK</em>

<em>T is absolute temperature: 273.15K</em>

<em>P is pressure: 1atm.</em>

<em />

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>
3 0
3 years ago
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