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yulyashka [42]
1 year ago
7

A sample of lead has the density of 58 g/mL and a mass of 15.2g, what is the volume of the lead?

Chemistry
1 answer:
torisob [31]1 year ago
4 0

Answer:

0.26 ml

Explanation:

d = m/V

=> V = m/d = 15.2/58 = 0.26 ml

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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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Write the balanced chemical equation for the combination (synthesis) reaction of magnesium with oxygen gas. phases are optional.
NISA [10]
The answer to this question would be:2 Mg<span> + O</span>2<span> = 2 MgO

Magnesium is a metal that can be oxidized easily. The reaction of magnesium and oxygen can be said as burning the magnesium. The reaction can provide a bright light with violet color. Magnesium used mostly in solid state and the oxygen is gas.</span>
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3 years ago
Balance the following chemical equations:
tamaranim1 [39]

Answer:

Explanation

=============

One

=============

Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + H2O

Focus on the NO3. This is an odd problem and you usually do not focus on the complex ion. But this one works easiest if you do.

The problem now is going to be the oxygens. There are 2 with the Calcium and only 1 free one going to the water. (The NO3 has been taken care of in the last step).

Ca(OH)2 + 2HNO3 -----> Ca(NO3)2 + 2H2O

Count the atoms. I think this equation is balanced.

atom                      Left              Right         Result

Ca                            1                    1               Balanced

O                              8                   8              Balanced

H                              2 + 2              2*2         Balanced

N                               2                    2            Balanced

===========

Two

===========

CH4 + O2====>  CO2 + H2O

Start with the hydrogens.

The right side requires a 2

CH4 + O2 ===>  CO2 + 2H2O

Now look at the oxygens. There are 4 on the right. and only 2 on the left. You need to multiply O2 by 2

CH4 + 2O2 ===>  CO2 + 2H2O

Each side has 1 Carbon 4 hydrogens and 4 oxygens. The equation is balanced.

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3 years ago
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