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Sladkaya [172]
3 years ago
6

How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?

Chemistry
1 answer:
max2010maxim [7]3 years ago
6 0

Answer: A volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

Explanation:

Given: V_{1} = ?,        M_{1} = 3.0 M\\

V_{2} = 100.0 mL,       M_{2} = 18 M

Formula used to calculate the volume is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL

Thus, we can conclude that a volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

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Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?

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How many grams of aluminum in required to produce 17.5 g of hydrogen
azamat

Mass of Aluminum = 157.5 g

<h3>Further explanation</h3>

Given

17.5 g of Hydrogen

Required

Mass of Aluminum

Solution

Reaction

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

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