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motikmotik
2 years ago
9

An aluminum cylinder has a radius of 3.2 cm and a length of 8.6 cm. What is the mass of the aluminum cylinder if aluminum has a

density of 2.7 g/cm3?
please help :)
Chemistry
1 answer:
Hatshy [7]2 years ago
5 0

Answer:

750g of Aluminum

Assuming 8.6 cm lenth is the height of the cylinder. The volume of a cylinder is: V = \pi*r^2*h

V = 3.14cm x 10.24cm x 3.6cm

V = 280cm^3

Now density = mass/volume

2.7g/cm^3 = mass/280cm^3

2.7g/cm^3 x 280cm^3 = mass/<u>280cm^3</u> x <u>280cm^3</u>

= 750g of Aluminum

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cricket20 [7]

Answer:

600 seconds, or 10 minutes

Explanation:

300m/ 0.5m = 600s

7 0
2 years ago
Calculate the molarity of the two solutions. The first solution contains 0.550 mol of NaOH in 1.40 L of solution.
olya-2409 [2.1K]

Answer:

0.393 mol/L.

Explanation:

The following data were obtained from the question:

Number of mole of NaOH = 0.550 mol

Volume of solution = 1.40 L

Molarity of NaOH =.?

Molarity of a solution is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the molarity of the NaOH solution as follow:

Number of mole of NaOH = 0.550 mol

Volume of solution = 1.40 L

Molarity of NaOH =.?

Molarity = mole / Volume

Molarity of NaOH = 0.55 / 1.4

Molarity of NaOH = 0.393 mol/L

Thus, the molarity of the NaOH solution is 0.393 mol/L.

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3 years ago
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Answer: This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
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3 years ago
If 4.0 g of helium gas occupies a volume of 22.4 L at 0 o C and a pressure of 1.0 atm, what volume does 3.0 g of He occupy under
WINSTONCH [101]

Answer:

the volume occupied by 3.0 g of the gas is 16.8 L.

Explanation:

Given;

initial reacting mass of the helium gas, m₁ = 4.0 g

volume occupied by the helium gas, V = 22.4 L

pressure of the gas, P = 1 .0 atm

temperature of the gas, T = 0⁰C = 273 K

atomic mass of helium gas, M = 4.0 g/mol

initial number of moles of the gas is calculated as follows;

n_1 = \frac{m_1}{M} \\\\n_1 = \frac{4}{4} = 1

The number of moles of the gas when the reacting mass is 3.0 g;

m₂ = 3.0 g

n_2 = \frac{m_2}{M} \\\\n_2 = \frac{3}{4} \\\\n_2 = 0.75 \ mol

The volume of the gas at 0.75 mol is determined using ideal gas law;

PV = nRT

PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\since, \ \frac{RT}{P} \ is \ constant,\  then;\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1n_2}{n_1} \\\\V_2 = \frac{22.4 \times 0.75}{1} \\\\V_2 = 16.8 \ L

Therefore, the volume occupied by 3.0 g of the gas is 16.8 L.

4 0
2 years ago
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