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3 years ago

How many moles of magnesium hydroxide, Mg(OH)2 can be created using 2.23 x 10^24 oxygen atoms?

Chemistry

1 answer:

igor_vitrenko [27]3 years ago

7 0

Answer:

32.07 g/mole.

Explanation:

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NaOH+HCl->............

Oxana [17]

Answer:

NaCl+H20

Explanation:

It is a neutralisation reaction in which NaOH is a base and HCl is an acid. On reaction it forms salt and water.

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2 years ago

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Which of the following is a unit of length? O A. A liter O B. A kilogram C. A meter O D. A degree

sergeinik [125]

Answer:

C.) A meter

Explanation:

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2 years ago

What conditions make G always positive?

Bess [88]

The answer is c ......

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2 years ago

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What is the mass of 0.73 moles of AgNO3?

love history [14]

Answer:

124 g (3 sig figs)

124.011 g (6 sig figs

Explanation:

Step 1: Calculate g/mol for AgNO₃

Ag - 107.868 g/mol

N - 14.01 g/mol

O - 16.00 g/mol

107.868 + 14.01 + 16.00(3) = 169.878 g/mol

Step 2: Multiply 0.73 moles by molar mass

0.73 mol (169.979 g/mol)

124 grams of AgNO₃

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3 years ago

You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol

oee [108]

Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

$0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol$

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

$6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH$

Step 4: Calculate the molarity of the acetic acid solution

$M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M$

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

$6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g$

Step 6: Calculate the percentage of acetic acid in the solution

$\frac{0.4186g}{25.00mL} \times 100\% = 1.674 \%(m/V)$

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3 years ago

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