Please help me :( How many atoms are present in 179.0 g of iridium? A. 6.464×10 B. 1.157×10 C. 5696×10 D

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How many atoms are present in 179.0 g of iridium?

A. 6.464×10
B. 1.157×10
C. 5696×10
D. 1078 x 10

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Verity 5 years 2021-07-24T00:27:34+00:00 1 Answers 515 views 0

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  1. Helga
    0
    2021-07-24T00:29:18+00:00
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    Answer: There are 5.696 \times 10^{23} atoms present in 179.0 g of iridium.

    Explanation:

    Given: Mass = 179.0 g

    Moles is the mass of a substance divided by its molar mass. So, moles of iridium (molar mass = 192.217 g/mol) is as follows.

    Moles = \frac{mass}{molar mass}\\= \frac{179.0 g}{192.217 g/mol}\\= 0.931 mol

    According to the mole concept, 1 mole of every substance contains 6.022 \times 10^{23} atoms.

    Therefore, atoms present in 0.931 moles are calculated as follows.

    0.931 mol \times 6.022 \times 10^{23} atoms/mol\\= 5.696 \times 10^{23} atoms

    Thus, we can conclude that there are 5.696 \times 10^{23} atoms present in 179.0 g of iridium.

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