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soldi70 [24.7K]
3 years ago
9

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

Chemistry
2 answers:
muminat3 years ago
4 0

Hello!

Determine the mass of 5.20 moles of C6H12 (gram-formula mass = 84.2 grams/mole).

We have the following data:

m (mass) = ? 

n (number of moles) = 5.20 moles

MM (Molar mass of C6H12) ≈ 84.2 g/mol

Now, let's find the mass, knowing that:

n = \dfrac{m}{MM}

5.20\:\:\diagup\!\!\!\!\!\!\!mol = \dfrac{m}{84.2\:g/\diagup\!\!\!\!\!\!\!mol}

m = 5.20*84.2

\boxed{\boxed{m = 437.84\:g}}\end{array}}\qquad\checkmark

_______________________

I Hope this helps, greetings ... Dexteright02! =)

Lapatulllka [165]3 years ago
4 0
Molar mass C₆H₁₂ = 84.2 g/mol

1 mole ---------- 84.2 g
5.20 moles ---- ?

Mass C₆H₁₂ = 5.20 * 84.2 / 1

Mass = 437.84 /1

= 437.84 g

hope this helps!

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Answer:

The answer is

<h2>155 g</h2>

Explanation:

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<h3>mass = Density × volume</h3>

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A sample of nitrogen gas is collected over water at temperature of 20.0˚C. What is the pressure of the nitrogen gas if atmospher
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P_N=0.987atm

Explanation:

Hello there!

In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

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4 0
3 years ago
NEED HELP ASAP NOT DIFFICULT
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Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
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If a 0.4681 g Mg strip reacts with 0.650 M HCl in a 139.3 mL flask at 25oC, what is the minimum volume (mL) of HCl needed to com
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The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,

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The number of moles of HCl that is needed for the reaction is calculated below.
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<em>Answer: 59.26 mL of HCl</em>
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