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trasher [3.6K]
2 years ago
14

How many moles of lithium are present in a sample that contains 2.45x10^87 formula units of Li2SO4?​

Chemistry
1 answer:
ddd [48]2 years ago
3 0

The answer is: 8.14·10⁶³ moles of lithium are present.

N(Li₂SO₄) = 2.45·10⁸⁷; number of formula units of lithium sulfate.

n(Li₂SO₄) = N(Li₂SO₄) ÷ Na.

n(Li₂SO₄) = 2.45·10⁸⁷ ÷ 6.022·10²³ 1/mol.

n(Li₂SO₄) = 4.07·10⁶³ mol; amount of lithium sulfate

In one molecule of lithium sulfate, there are two atoms of lithium.

n(Li₂SO₄) : n(Li) = 1 : 2.

n(Li) = 2 · 4.07·10⁶³ mol.

n(Li) = 8.14·10⁶³ mol; amount of lithium atoms.

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Question 2 The metal molybdenum becomes superconducting at temperatures below 0.90K. Calculate the temperature at which molybden
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Answer:

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As given, molybdenum becomes superconducting at temperatures below 0.90 K.

Temperature in Kelvins can be converted in °C by relation:

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Temperature at which molybdenum becomes superconducting is -272.25 °C

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How many liters of 3.0 M NaOH solution will react with 2.4 mol H2SO4? (Remember to balance the equation.)
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Answer:

1.6 L is the volume of NaOH that has reacted

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We propose this rule of three:

1 mol of sulfuric acid needs 2 mol of NaOH to react to react

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