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3 years ago

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What do solids and liquids have in common?

1 answer:

m_a_m_a [10]3 years ago

7 0

They have a fixed volume.

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Are these two substances the same 1.96g/ml and 2.00g/ml

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How many milliliters of 2.19 M H2SO4 are required to react with 4.75 g of solid containing 21.6 wt% Ba(NO3)2 if the reaction is

Setler [38]

Answer:

1.7927 mL

Explanation:

The mass of solid taken = 4.75 g

This solid contains 21.6 wt% $Ba(NO_3)_2$ , thus,

Mass of $Ba(NO_3)_2$ = $\frac {21.6}{100}\times 4.75\ g$ = 1.026 g

Molar mass of $Ba(NO_3)_2$ = 261.337 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.026\ g}{261.337\ g/mol}$

$Moles= 0.003926\ mol$

Considering the reaction as:

$Ba(NO_3)_2+H_2SO_4\rightarrow BaSO_4+2HNO_3$

1 moles of $Ba(NO_3)_2$ react with 1 mole of $H_2SO_4$

Thus,

0.003926 mole of $Ba(NO_3)_2$ react with 0.003926 mole of $H_2SO_4$

Moles of $H_2SO_4$ = 0.003926 mole

Also, considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Molarity = 2.19 M

So,

$2.19=\frac{0.003926}{Volume\ of\ the\ solution(L)}$

Volume = 0.0017927 L

Also, 1 L = 1000 mL

<u>So, volume = 1.7927 mL</u>

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2 years ago

In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i

GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: $V_{1}$ = 21.35 mL, $M_{1}$ = 0.150 M

$V_{2}$ = 25.0 mL, $M_{2}$ = ?

Formula used to calculate molarity of $H_{2}SO_{4}$ is as follows.

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

Substitute the values into above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.15 M \times 21.35 mL = M_{2} \times 25.0 mL\\M_{2} = 0.1281 M$

As molarity is the number of moles of a substance present in a liter of solution.

Total volume of solution = $V_{1} + V_{2}$

= 21.35 mL + 25.0 mL

= 46.36 mL (1 mL = 0.001 L)

= 0.04636 L

Therefore, moles of acid required are calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\0.1281 M = \frac{no. of moles}{0.04635 L}\\no. of moles = 0.006 mol$

Thus, we can conclude that there are 0.006 moles of acid in the flask.

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2 years ago