If 25.0 mL of 0.100 M Ca(OH)2 is titrated with 0.200 M HNO3, what volume of nitric acid is required to neutralize the base?
1 answer:
Answer:
25 mL
Explanation:
The reaction that takes place is:
- Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O
First we<u> calculate how many Ca(OH)₂ moles</u> were spent in the titration:
- 25.0 mL * 0.100 M = 2.5 mmol Ca(OH)₂
Then we <u>convert Ca(OH)₂ moles into HNO₃ moles</u>, using the <em>stoichiometric ratio</em>:
- 2.5 mmol Ca(OH)₂ *
= 5.0 mmol HNO₃
Finally we <u>calculate the volume of required nitric acid solution</u>, using the <em>concentration</em>:
- 5.0 mmol ÷ 0.200 mmol/mL = 25 mL
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2CO + 2NO → 2CO₂ + N₂
Looking closely at the options given, several of them could be eliminated immediately since they are not balanced equations like B, C and E. Therefore, only A or D could be the answers.
The answer is option <span>D.) 2CO + 2NO → 2CO2 + N2
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The red atoms refer to oxygen atoms, the grey atoms are carbon atoms and the blue atoms are nitrogen. Therefore, the equation from the image suggests:
2CO + 2NO → 2CO₂ + N₂
Looking closely at the options given, several of them could be eliminated immediately since they are not balanced equations like B, C and E. Therefore, only A or D could be the answers.
The answer is option <span>D.) 2CO + 2NO → 2CO2 + N2
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<h2>Density = 0.75 g/mL</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 450 g
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Substitute the values into the above formula and solve for the density
That's
We have the final answer as
<h3>Density = 0.75 g/mL</h3>Hope this helps you