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stich3 [128]
3 years ago
11

How many moles of nitric acid are present in 35.0 ml of a 2.20 M solution?

Chemistry
1 answer:
vodomira [7]3 years ago
8 0

Answer:

There are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

Explanation:

Molarity of the solution = 2.20 M

Molarity=\frac{number\:of\:moles}{Volume\:of\:Solution\:in\:L}\\\\Number\:of\:moles=Molarity\times(Volume\:of\:Solution\:in\:L)\\\\Volume\:of\:Solution=35\:mL=35\times10^{-3}L\\\\Number\:of\:moles=2.20\times35\times10^{-3}=77\times10^{-3}\:moles\:of\:HNO_{3}

Therefore, there are 77 millimoles of nitric acid present in 35.0 mL of a 2.20 M solution

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2 years ago
How many moles are in 5.3 X 10^6 atoms of Calcium (Ca)?
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Answer:

8.801×10^-18 moles Ca

Explanation:

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3 years ago
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Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

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where,

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Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

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Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

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4 years ago
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