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s344n2d4d5 [400]
3 years ago
11

__Co + __F2 --> __CoF3

Chemistry
1 answer:
Natalka [10]3 years ago
8 0

Answer:

2Co + 3F2 --> 2CoF3

Explanation:

When balancing an equation we need to ensure that the products used equal the reactants produced. Hence to balance the equation we need 3 F2 molecules so we have 6F atoms in total. and since 1 Co molecule bonds to 3 F molecules. We are gonna need to 2 Co molecules.

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Answer:

C.

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3 0
3 years ago
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What mass of potassium bromide (in grams) do you need to make 250.0 mL of a 1.50 M potassium bromide solution?
Vedmedyk [2.9K]

Answer:

44.63g

Explanation:

First, let us calculate the number of mole of KBr in 1.50M KBr solution.

This is illustrated below:

Data obtained from the question include:

Volume of solution = 250mL = 250/1000 = 0.25L

Molarity of solution = 1.50M

Mole of solute (KBr) =.?

Molarity is simply mole of solute per unit litre of solution

Molarity = mole /Volume

Mole = Molarity x Volume

Mole of solute (KBr) = 1.50 x 0.25

Mole of solute (KBr) = 0.375 mole

Now, we calculate the mass of KBr needed to make the solution as follow:

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Mole of KBr = 0.375 mole

Mass of KBr =?

Mass = number of mole x molar Mass

Mass of KBr = 0.375 x 119

Mass of KBr = 44.63g

Therefore, 44.63g of KBr is needed to make 250.0mL of 1.50 M potassium bromide (KBr) solution

8 0
3 years ago
Smoking marijuana can be as unhealthy as smoking cigarettes and ___________
Afina-wow [57]
The answer is all of them. But in a school situation, A would be appropriate. This is because the smoke gets into the smokers lungs, whether it be marijuana or tobacco, and can cause cancer cells to grow in your lungs and enter your bloodstream.
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3 0
3 years ago
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Which describes the volume of 1 mol of gas at standard temperature and pressure?
densk [106]

Answer:

B) The volume is the same for any gas.

Explanation:

Considering the ideal gas equation as:-

PV=nRT

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V is the volume

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T is the temperature  

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Pressure = 1 atm  

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Thus, correct option is:- B) The volume is the same for any gas.

6 0
3 years ago
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