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krek1111 [17]
3 years ago
9

A solution was prepared by mixing 50.00 ml of 0.100 M of HNO3 and 100.00 ml of 0.200 M HNO3. Calculate the molarity of the final

solution of nitric acid
Chemistry
1 answer:
valentina_108 [34]3 years ago
4 0

Answer:

The molarity of this final solution is 0.167 M

Explanation:

Step 1: Data given

Volume of a 0.100 M HNO3 solution = 50.0 mL

Volume of a 0.200 M HNO3 = 100.0 mL

Step 2: Calculate moles

The final molarity must lie between 0.1M and 0.2M  

Moles = molarity * volume

Moles HNO3 in 50mL of a 0.100M solution = 0.05 L *0.100 M = 0.005 mol

Moles HNO3 in 100mL of a 0.200M solution = 0.100 L*0.200 = 0.020mol

total moles = 0.005+0.020 = 0.025 moles in 150mL solution = 0.150L

Step 3: Calculate molarity of final solution

Molarity = mol / volume

Molarity 0.025 moles /0.150  L

Molarity = 0.167M

The molarity of this final solution is 0.167 M

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Please help on this one?
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Answer:

\text{C. } _{36}^{85}\text{Kr}

Explanation:

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_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} +\, _{x}^{y}\text{X}

The main point to remember in balancing nuclear equations is that

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