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Anton [14]
3 years ago
6

How many grams of CO2 are in 2.1 mol of the compound? a) 21.0g c) 66.0g

Chemistry
1 answer:
Lilit [14]3 years ago
8 0
1 mole CO2 = 44.0096 grams CO2


<span>2.1 mol CO2 x (44.0096 grams CO2/1 mole CO2) = 92.4 grams CO2</span>
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Naya [18.7K]

Answer:

V_{base}=10.2mL

Explanation:

Hello.

In this case, since the neutralization of the acid requires equal number of moles of both acid and base:

n_{acid}=n_{base}

Whereas we can express it in terms of concentrations and volumes:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we can compute the volume of sodium hydroxide (base) as follows:

V_{base}=\frac{M_{acid}V_{acid}}{M_{base}} \\\\V_{base}=\frac{87.0mL*0.234M}{2.00M}\\ \\V_{base}=10.2mL

Best regards.

8 0
3 years ago
Which of the following is a pure substance? <br> brass<br> sodium<br> rocks<br> steel
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Homogeneous mixture<span>steel

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3 0
4 years ago
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3. A thin lead apron is used to protect patients from harmful X rays. If the sheet measures 75.0 cm by 55.0 cm by 0.10 cm, and t
DanielleElmas [232]

Answer:

4.67 kg

Explanation:

Given data

  • Dimensions of the lead sheet: 75.0 cm by 55.0 cm by 0.10 cm
  • Density of lead: 11.3 g/cm³

Step 1: Calculate the volume of the sheet

The volume of the sheet is equal to the product of its dimensions.

V = 75.0 cm \times  55.0 cm \times 0.10 cm = 413 cm^{3}

Step 2: Calculate the mass of the sheet

The density (ρ) is equal to the mass divided by the volume.

\rho = \frac{m}{V} \\m = \rho \times V = \frac{11.3g}{cm^{3} }  \times 413cm^{3} = 4.67 \times 10^{3} g = 4.67 kg

6 0
3 years ago
What is the concentration of each ion in a solution that is prepared by dissolving 5.00 g of ammonium chloride in enough water t
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Answer:

C = 0.08M

Explanation:

molar mass of AlCl3

Al =27

Cl = 35.5

27+3(35.5) =133.5g/mol

n= mass/Molar mass

n =CV

CV = mass/molar mass

C x 500 x 10^-³ = 5/133.5

C x 500 x 10^-³ = 0.04

C = 0.04/500 x 10^-³

C = 0.08M

4 0
3 years ago
Which of the following lists the given objects in the correct mass order from lowest to highest?
Mrrafil [7]
Sorry no se inglés
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