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Alex777 [14]
3 years ago
11

How many grams are in 1.11 moles of manganese sulfate, Mn3(SO4)7?

Chemistry
2 answers:
UNO [17]3 years ago
8 0
The correct answer is 167.60889
UNO [17]3 years ago
6 0

Answer:

There are 167.60889 grams in 1.11 moles of manganese sulfate. The molecular formula of manganese sulfate is MnSO4.

Explanation:

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7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

<h3>How to find the number of moles ?</h3>

Number of moles = \frac{\text{Given number of atoms}}{\text{Avogadro's Number}}

     

<h3>What is Avogadro's Number ?</h3>

Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's Constant / Avogadro's Number.

Avogadro's Number = 6.022 × 10²³

Now put the values in above formula we get

Number of moles = \frac{\text{Given number of atoms}}{\text{Avogadro's Number}}

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Thus from the above conclusion we can say that 7.32 moles of Chromium is present in 4.41 × 10²⁴ atoms.

Learn more about the Avogadro's Number here: brainly.com/question/1581342

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I think you wanted to ask what is the meaning by the term alloy:

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