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andrey2020 [161]

2 years ago

7

Exactly 25.0 mL of an aqueous solution of barium hydroxide required 32.0 mL of 0.200 M nitric acid to neutralize it. What is the

concentration of the barium hydroxide solution?
Give only your numerical answer and do not include the M. For example, 0.262 or 1.25.

2 answers:

Brrunno [24]2 years ago

7 0

Answer:

2000m the example 25.0\0.200

notka56 [123]2 years ago

6 0

Answer:

$\frac{0.200 \times 320}{57} = .112$

Hope this helps

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What are the three particles of an atom?

Dmitry [639]

The three particles found in an atom are the protons, neutrons and electrons. Protons have a positive charge. Electrons have negative charge. Lastly, neutrons have no net electrical charge. Protons and neutrons are much heavier than electrons and are located in the center of the atom.

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2 years ago

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 M AgNO3 requires 20.22 mL o

Nina [5.8K]

Answer:

The concentration of NaCl = 0.3374 M

Explanation:

Given :

Molarity of AgNO₃ = 0.2503 M

Volume of AgNO₃ = 20.22 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 20.22×10⁻³ L

Molarity of a solution is the number of moles of solute present in 1 L of the solution.

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

The formula can be written for the calculation of moles as:

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

Thus,

$Moles\ of\ AgNO_3 =Molarity \times {Volume\ of\ the\ solution}$

$Moles\ of\ AgNO_3 =0.2503 \times {20.22\times 10^{-3}}\ moles$

$Moles\ of\ AgNO_3 = 5.0611 \times 10^{-3} moles$

The chemical reaction taking place:

$AgNO_3_(aq) + NaCl_(aq) \rightarrow AgCl_(s) + NaNO_3_(aq)$

According to reaction stoichiometry:

<u>1 mole</u> of AgNO₃ reacts with <u>1 mole</u> of NaCl

Thus,

5.0611×10⁻³ moles of AgNO₃ reacts with 5.0611×10⁻³ moles of NaCl

Thus, moles of NaCl required = 5.0611×10⁻³ moles

Volume of NaCl required = 15.00 mL

The conversion of mL into L is shown below:

$1 mL= 10^{-3} L$

Thus, volume of the solution = 15.00×10⁻³ L

Applying in the formula of molarity as:

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

$Molarity\ of\ NaCl=\frac{5.0611\times 10^{-3}}{15.00\times 10^{-3}}$

$Molarity\ of\ NaCl= 0.3374 M$

<u>Thus, the concentration of NaCl = 0.3374 M</u>

6 0

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Elena L [17]

After careful consideration your answer is...

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~Alanna~

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2 years ago

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