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nataly862011 [7]
3 years ago
10

Can someone please help me on the second quick Check!!!!

Chemistry
2 answers:
alexgriva [62]3 years ago
8 0
#1 fossils paragraph 2
Nezavi [6.7K]3 years ago
4 0
The organisms would have not been able to travel
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Calculate the number of moles and the mass of the solute in each of the following solutions:
frosja888 [35]

<u>Answer:</u>

<u>For a:</u> The number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

<u>For b:</u> The number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

<u>For c:</u> The number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

<u>For d:</u> The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}    .....(1)

To calculate the number of moles of a substance, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(2)

  • <u>For a:</u>

Molarity of KI = 8.23\times 10^{-5}M

Volume of solution = 325 mL = 0.325 L     (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol

Now, using equation 2, we get:

Moles of KI = 2.7\times 10^{-5}mol

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g

Hence, the number of moles of KI are 2.7\times 10^{-5} and mass is 4.482\times 10^{-3}g

  • <u>For b:</u>

Molarity of sulfuric acid = 22\times 10^{-5}M

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol

Now, using equation 2, we get:

Moles of sulfuric acid = 1.65\times 10^{-5}mol

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g

Hence, the number of moles of sulfuric acid are 1.65\times 10^{-5} and mass is 1.617\times 10^{-3}g

  • <u>For c:</u>

Molarity of potassium chromate = 0.1135M

Volume of solution = 0.250 L

Putting values in equation 1, we get:

0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol

Now, using equation 2, we get:

Moles of potassium chromate = 2.84\times 10^{-2}mol

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g

Hence, the number of moles of potassium chromate are 2.84\times 10^{-2} and mass is 5.51 g.

  • <u>For d:</u>

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

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3 years ago
Zn(s)+S(s)-ZnS(s)<br> What is the product and physical state
Yuki888 [10]
<span> The Reactants are Zinc (Zn) and Sulfur (S).
The Product is Zinc Sulfide (ZnS).
All of them are solids.
The combined masses of the reactants must be 14 grams, too. Later in Chemistry you'll learn that's not really true, but it is for now.
Hope This Helps:)
</span>
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Scientists use scientific notation to represent very small or very large numbers because this notation increases the
Tems11 [23]
...because this notation increases the convenience in using the numbers 
3 0
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Uranium-234 is unstable and undergoes two subsequent alpha emissions. What is the resulting nuclide from this transformation?
MArishka [77]

Answer:Radium

Explanation:

The nuclear reaction involving two alpha emissions of 234 U is shown in the diagram. This leads to the formation of a 226Ra nucleus.

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What are the primary colors?
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The primary colors are blue, red, and yellow.
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