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IgorC [24]
3 years ago
10

8 moles Cl to grams 3

Chemistry
1 answer:
kirill [66]3 years ago
7 0

Answer: I don’t understand what your asking elaborate more

Explanation:

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Which of the following is an ethical question?
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Answer:

will cloning technology be useful?

Explanation:

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3 years ago
The relationship between pressure and volume, when moles and temperature of a gas are held constant, is: PV = k. We could say th
pochemuha
You can say that if the volume of the gas is halved, the pressure is doubled.

The expression shows that pressure and volume are inversely proportional if temperature and amount of gas is held constant.  This means that if volume goes down the pressure needs to go up.  That also means that in order to maintain the K value, if pressure is doubled the volume needs to be halved and if the pressure is halved the volume needs to be doubled.

This relationship only works if we assume everything else (Temperature and moles of gas) to be constant.
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4 years ago
Which Atom is most likely to form a metallic bond
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“Aluminum” is most likely to
8 0
3 years ago
Read 2 more answers
To make a 2.0-mole solution, how many moles of solute must be dissolved in 0.50 liters of solution?
Flauer [41]

There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. The concentration of the solution is calculated as follows:

 Molarity = 2.0 mole / L solution

<span>2.0 mole / L solution ( 0.50 Liters ) = 1 mole solute</span>

<span>The correct answer is the third option. One mole of solute needed to make 0.50 liters of 2M solution.</span>

3 0
3 years ago
Calculate the temperature of a 0.50 mol sample of a gas at 0.987 atm and a volume of 12 L.
Aleonysh [2.5K]

Answer: 15.5^0C

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = 0.987 atm

V = Volume of gas = 12 L

n = number of moles = 0.50

R = gas constant =0.0821Latm/Kmol

T =temperature =  ?

T=\frac{PV}{nR}

T=\frac{0.987atm\times 12L}{0.0820 L atm/K mol\times 0.50mol}=288.5K=(288.5-273)^0C=15.5^0C

Thus the temperature of a 0.50 mol sample of a gas at 0.987 atm and a volume of 12 L is 15.5^0C

8 0
3 years ago
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