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3 years ago

When people respond to questions saying the answer is in an attached link, is it safe?

Chemistry

1 answer:

Mashcka [7]3 years ago

7 0

Answer:

Personally, I believe it is unsafe. When people include a link in their answer, they are breaking the Brainly code, which states that a link to an unknown website is prohibited material.

Explanation:

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B stratospheric zone

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3 years ago

The most penetrating neclear radiation is

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Radioactive is the most penetrating nuclear radiation

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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

3 years ago

6) A student measures out 96.21 g of sulfur for an experiment. How many moles of Sulfur are in this

barxatty [35]

Atomic mass of Sulfur = 32g

32g of Sulfur is one mole.

1g of Sulfur is $\frac{1}{32} moles$

96.21g of Sulfur is $\frac{96.21}{32} moles=> 3moles(appx)$

5 0

3 years ago

Read 2 more answers

How many Joules must be removed to condense 150.0g of steam at 100.0 degrees Celsius to water at 100.0 degrees celcius

Elina [12.6K]

Answer:

339kJ

Explanation:

Given parameters:

Mass of steam = 150g = 0.15kg

Initial temperature of steam = 100°C

Final temperature of water = 100°C

Unknown:

Quantity of heat that must be removed to condense the steam = ?

Solution:

The heat involved here is a latent heat because there is no change temperature. The process is just a phase change.

H = mL

m is the mass

L is the latent heat of vaporization = 2,260 kJ/kg

Insert the parameters and solve;

H = 0.15kg x 2,260 kJ/kg

H = 339kJ

7 0

3 years ago