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3 years ago

A dozen is a counting unit that refers to 12 and a pair is a counting unit that refers to

Chemistry

1 answer:

seraphim [82]3 years ago

8 0

Answer: 1 mol contains 6.02214076×1023 of the specified particles

Explanation:

Usually its easier to work with moles using grams of what you are counting for example: 1 mole of carbon is equal to 12 grams 1 mole of oxygen is equal to approximately 16 grams and so on.

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All of the following describe work being done except:

SIZIF [17.4K]

Answer:

A. standing in line to pay for groceries.

Explanation:

Just standing there is not doing work, All of the other options are examples of work.

7 0

4 years ago

How do sulfur and phosphorus move from the biotic to the abiotic pools during their cycle?

cluponka [151]

There are different ways, but the most common is through water.Hope this helps.

4 0

3 years ago

How many milliliter of a solution of 4.00KI are needed to prepare 250.0mL of 0.760 KI

Alexeev081 [22]

Answer:

47.5 mL

Solving:

M1 = 4.00 M

V1 = ?

M2 = 0.760 M

V2 = 0.250 L

---

M1 * V1 = M2 * V2

V1 = ( M2 * V2 ) / M1

V1 = ( 0.760 * 0.250 ) / 4.00

V1 = ( 0.190 ) / 4.00

V1 = 0.0475 L

3 0

4 years ago

Calculate the number of moles and the mass of the solute in each of the following solutions:

frosja888 [35]

Answer:

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

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

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

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

Explanation:

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)

For a:

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 $4.482\times 10^{-3}g$

For b:

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 $1.617\times 10^{-3}g$

For c:

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.

For d:

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.

3 0

4 years ago

The tarnish that forms on objects made of silver is solid silver sulphide. This can be removed by reacting it with aluminium met

steposvetlana [31]

Answer:

5.83 mol.

Explanation:

From the balanced reaction:

2Al + 3Ag₂S → 6Ag + Al₂S₃,

It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.

Al reacts with Ag₂S with (2: 3) molar ratio.

So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.

The limiting reactant is Ag₂S.

The excess "left over" reactant is Al.

The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.

6 0

3 years ago