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Flura [38]
3 years ago
12

1. An unknown, rectangular substance measures 3.60 cm high, 4.21 cm long, and 1.17 cm wide. If the mass is 21.3 g, what is this

substance’s density (in grams per milliliter)?
Chemistry
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

1.20g/mL

Explanation:

Step 1:

Data obtained from the question. This include the following:

Height of substance = 3.60 cm

Length of substance = 4.21 cm

Width of substance = 1.17 cm

Mass of substance = 21.3 g

Density of substance =...?

Step 2:

Determination of the volume of the substance.

The volume of the substance can be obtained as follow:

Volume = length x width x height

Volume = 4.21 x 1.17 x 3.60

Volume = 17.73cm³

Recall:

1cm³ = 1mL

Therefore, 17.73cm³ = 17.73mL

Therefore, the volume of the substance is 17.73mL.

Step 3:

Determination of the density of the substance.

The Density of a substance is simply defined as the mass per unit volume of the substance. Mathematically, it is represented as:

Density = Mass /volume

With the above formula, we can obtain the density of the substance as follow:

Mass of substance = 21.3 g

Volume of substance = 17.73mL

Density =..?

Density = Mass /volume

Density = 21.3/17.73

Density = 1.20g/mL

Therefore, the density of the substance is 1.20g/mL.

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Question 2(Multiple Choice Worth 5 points) (06.04 HC) During a laboratory experiment, 36.12 grams of Al2O3 was formed when O2 re
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Answer:

8.70 liters.

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>3O₂ + 4Al → 2Al₂O₃,</em>

It is clear that 3.0 moles of O₂ react with 4.0 moles of Al to produce 2.0 Al₂O₃.

  • Firstly, we need to calculate the no. of moles (n) of 36.12 g of Al₂O₃:

n = mass/molar mass = (36.12 g)/(101.96 g/mol) = 0.3543 mol.

<u><em>using cross multiplication:</em></u>

3.0 mol of O₂ produces → 2.0 mol of Al₂O₃.

??? mol of O₂ produces → 0.3543 mol of Al₂O₃.

∴ The no. of moles of O₂ needed to produce 36.12 grams of Al₂O₃ = (3.0 mol)(0.3543 mol)/(2.0 mol) = 0.5314 mol.

  • Now, we can find the volume of O₂ used during the experiment:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 1.4 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 0.5314 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 280 K).

∴ V = nRT/P = (0.5314 mol)(0.0821 L.atm/mol.K)(280 K)/(1.4 atm) = 8.726 L ≅ 8.70 L.

<em>So, the right choice is: 8.70 liters.</em>

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