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Ainat [17]
3 years ago
6

A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? U

se .
Chemistry
2 answers:
umka2103 [35]3 years ago
7 0

Answer:

option A

Explanation:

just took test on e2020

Aleks [24]3 years ago
6 0
Hello!

The concentration of the final solution when a<span> chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water is 0,3 M

To calculate that, you'll need to use the dilution law, where initial and final concentrations are M1 and M2 respectively, and initial and final volumes are V1 and V2, as shown below. Keep in mind that the final volume is the sum of the 200 mL of water and the 50 mL of H</span>₂SO₄ that were added by the teacher. 

M2= \frac{M1*V1}{V2}= \frac{1,50 (mol H_2SO_4/L)*50mL}{(50 mL + 200 mL)}=0,3(mol H_2SO_4/L)

Have a nice day!
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For each of the following compounds, identify what type of bonding holds them together.
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Explanation

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Calculate the molarity of a solution consisting of 25.0 g of KOH in 3.00 L of solution.
melomori [17]

Answer:

0.15M

Explanation:

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You need to calculate the number of moles from the given grams. The molar mass of KOH is (39.098+ 16 +1.008)= 56.106g. To calculate the mols of KOH, \frac{25.0g}{1} × \frac{1 mol}{56.106g} = 0.44558... mol, you see that the grams unit cancel out leaving you with mol as the unit.

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ankoles [38]

Answer:

When nitric acid combine with sodium hydroxide the salt formed is called sodium nitrate. option B

Explanation:

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HNO₃(aq) + NaOH(aq)       →      NaNO₃(aq) + H₂O(l)

Ionic equation:

H⁺NO₃⁻(aq) + Na⁺OH⁻(aq)       →      Na⁺NO₃⁻(aq) + H₂O(l)

Net ionic equation:

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The Na⁺(aq) and NO₃⁻(aq) are spectator ions that's why these are not written in net ionic equation. The water can not be splitted into ions because it is present in liquid form.

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