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dmitriy555 [2]
3 years ago
14

How many grams of CO2 are in 2.1 mol of the compound?

Chemistry
2 answers:
Komok [63]3 years ago
6 0

the awnser is d hope i helped

lawyer [7]3 years ago
3 0
The molar mass of a substance is the mass of one mole of that substance. Therefore, in order to calculate the mass that is present while the number of moles are known, we simply multiply the number of moles present by the molar mass. In the case of carbon dioxide, the molar mass is obtained by adding the mass of one carbon atom (12) and two oxygen atoms (32). The molar mass is 12 + 32 = 44. Therefore, the mass of carbon dioxide is 2.1 * 44 = 92.4 grams. The answer is D<span>.</span>
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Zn + 2HCI --&gt; ZnCl2 + H2
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Moles= mass\ relative formula mass(Ar)
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2 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

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Answer:

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