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3 years ago

Calculate a mass of 12.044×10^23 of calcium carbonate

Chemistry

1 answer:

tigry1 [53]3 years ago

4 0

0.02 g of calcium carbonate (CaCO₃)

Explanation:

To solve the question we use the Avogadro's number using the following reasoning:

if 1 moles of calcium carbonate contains 6.022 × 10²³ units of calcium carbonate

then X moles of calcium carbonate contains 12.044 × 10²³ units of calcium carbonate

X = (1 × 12.044 × 10²³) / 6.022 × 10²³ = 2 moles of calcium carbonate

mass = number of moles × molecular weight

mass of calcium carbonate (CaCO₃) = 2 / 100 = 0.02 g

Learn more about:

Avogadro's number

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Answer:

K = 0.167 s⁻¹

Explanation:

1) Rate law, at a given temperature:

Since all the data are obtained at the same temperature, the equilibrium constant is the same.

Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:

r = K [A]ᵃ [B]ᵇ

2) Use the data from the table

Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s

Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0

Use the first and second set of data to find the exponent a:

r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s

r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s

Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]

2ᵃ = 2 ⇒ a = 1

3) Write the rate law

r = K [A]¹ [B]⁰ = K[A]

This means, that the rate is independent of reactant B and is of first order respect reactant A.

4) Use any set of data to find K

With the first set of data

r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹

Result: the rate constant is K = 0.167 s⁻¹

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4 years ago

What element has three electrons in its 5d sublevel?

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3 years ago

Which one of the following statements is not true concerning 2.00 L of a 0.100 M solution of Ca3(PO4)2?

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Answer: The correct answer is Option B.

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)

For A:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{2.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 2.00L)=0.200mol$

Moles of calcium phosphate = 0.200 moles

For B:

1 mole of calcium phosphate contains 3 moles of calcium atoms, 2 moles of phosphate atoms and 8 moles of oxygen atoms.

So, 0.200 moles of calcium phosphate will contain = $(8\times 0.200)=1.6$ moles of oxygen atoms.

Moles of oxygen atoms = 1.6 moles

For C:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 1.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{1.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 1.00L)=0.100mol$

Moles of calcium ions = $(0.100\times 3)=0.300$ moles

For D:

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 5.00 L

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{5.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 5.00L)=0.500mol$

Moles of phosphorus atoms = $(0.500\times 2)=1.00$ moles

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of atoms

Number of phosphorus atoms in 0.500 moles of calcium phosphate will be = $(1.00\times 6.022\times 10^{23})=6.022\times 10^{23}$

For E:

1 mole of calcium phosphate contains 3 moles of calcium ions and 2 moles of phosphate ions.

So, 0.200 moles of calcium phosphate will contain = $(3\times 0.200)=0.600$ moles of calcium ions

Moles of calcium ions = 0.600 moles

Hence, the correct answer is Option B.

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3 years ago

Calculate a mass of 12.044×10^23 of calcium carbonate​

Calculate a mass of 12.044×10^23 of calcium carbonate