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Serga [27]
3 years ago
11

At what temperature will 0.0705 mol of Cl2 exert a pressure of 900. torr at a volume of 0.750L?

Chemistry
1 answer:
kompoz [17]3 years ago
8 0

Answer:

It will exert the pressure at a temperature of <em><u>153.44 K</u></em>

Explanation:

To answer this question, we shall be using the ideal gas equation;

PV = nRT

Since we are calculating the temperature, it can be made the subject of the formula.

Thus, this can be T = PV/nR

where P is the pressure = 900 torr

V is the volume = 0.75 L

n is the number of moles = 0.0705 mol

R is the molar gas constant = 62.4 L.Torr.k^{-1}.mol^{-1}

Plugging these values into the equation, we have;

T = (900 × 0.75)/(0.0705 × 62.4)

T = 153.44 K

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Answers are:

2. It pushes on all objects that are on Earth’s surface.

3. It can be measured in atmospheres or kilopascals.

Barometric pressure (atmospheric pressure), is the pressure within the atmosphere of Earth

Atmospheric pressure decreases with increasing height, because there are fewer air molecules above a given object.

Barometer is an instrument used in meteorology to measure atmospheric pressure.

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4 0
3 years ago
Luciana explains to Eric how electricity flows in a circuit. Which sentences are a part of her explanation? Select all that appl
Flauer [41]
I’m pretty sure it c hope that helped
8 0
3 years ago
Read 2 more answers
1. How much taller is the average person today than the average person 100 years ago?
andreev551 [17]
5 inches I am not sure but I THINK it’s 5 inches sry if I’m wrong
4 0
3 years ago
What does chemical reaction describe
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A chemical reaction (signs)

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3 0
3 years ago
Under the right conditions aluminum will react with chlorine to produce aluminum chloride.
salantis [7]

Answer:

m_{Al}=9.42gAl

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2}  *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl

Regards!

8 0
3 years ago
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