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3 years ago

Convert 1.5x10^25 atoms of hydrogen into moles of Hydrogen

Chemistry

2 answers:

777dan777 [17]3 years ago

8 0

Moles = # of atoms / 6.023 x 10^23 (Avogadro’s constant)

1.5 x 10^23 / 6.023 x 10^23 = 51.5 moles of Hydrogen

allsm [11]3 years ago

7 0

Sorry man I need some points but I do t even know the answer

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How many moles of KBr will be produced from 10.51 moles of BaBr2?

gregori [183]

Answer:

21.02moles of KBr

Explanation:

Parameters given:

Number of moles BaBr₂ = 10.51moles

Complete reaction equation:

BaBr₂ + K₂SO₄ → KBr + BaSO₄

Upon inspecting the given equation, we find out that the atoms are not balanced on both sides of the equation:

The balanced equation is:

BaBr₂ + K₂SO₄ → 2KBr + BaSO₄

From the equation:

1 mole of BaBr₂ produces 2 moles of KBr

∴ 10.51 moles of BaBr₂ will yield (2 x 10.51) moles = 21.02moles of KBr

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4 years ago

What is the name of Bel on the periodic table

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3 years ago

5. Mr. Martin's Spanish class is 45 minutes long. If it starts at 3:30, what time does it end?

zubka84 [21]

It would be 4:15 . Just add 3:30 plus 45 too get the answer

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3 years ago

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6. Would you describe each of these temperatures as warm, hot, or cold?

mash [69]

Answer:

b: Hot

a: Cold

c: Cold

d: Warm

e: Warm

f: Cold

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8 0

3 years ago

A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t

andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

$OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)$

So, moles of product formed are calculated as follows.

$\frac{3}{2} \times 0.17 mol \\= 0.255 mol$

Hence, the given data is as follows.

$n_{1}$ = 0.17 mol, $n_{2}$ = 0.255 mol

$V_{1}$ = 19.5 L, $V_{2} = ?$

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$

Substitute the values into above formula as follows.

$\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L$

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0

3 years ago