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3 years ago

What is the mass of calcium in 2.3•10^23 molecules of calcium phosphate

Chemistry

1 answer:

Katena32 [7]3 years ago

3 0

Explanation:

N(Ca)=3×N(Ca₃(PO₄)₂)=

=3×2.3×10²³=6.9×10²³

6.02×10²³→40g

6.9×10²³→Xg

m(Ca)=X≈45.85g

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Which statement is true of a reversible reaction at equilibrium?

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Explanation:

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A chemical reaction produces 3.81g potassium chloride (KCl). How many formula units of potassium chloride are there? *Find Molar

Aleks04 [339]

Answer:

The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²

Explanation:

The given parameters is as follows;

The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g

The required information = The number of formula units of potassium chloride (KCl)

The Molar Mass of KCl = 74.5513 g/mol

$The \ number \ of \ moles \ of \ a \ substance, n = \dfrac{The \ mass \ of \ the substance}{The \ Molar \ Mass \ of \ the \ substance}$

Therefore, we have;

$The \ number \ of \ moles \ of \ KCl= \dfrac{3.81 \ g}{74.5513 \ g/mol} \approx 0.051106 \ moles$

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0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³ ≈ 3.077588 × 10²² formula units

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2 years ago

WS Percent yield don’t understand how to do would appreciate the help

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Answer:

1. Theoretical yield of NaOH is 22.72 g

2. Percentage yield of NaOH = 22.14%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

NaHCO₃ —> NaOH + CO₂

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 1 mole (i.e 40 g) of NaOH and 1 mole (i.e 44.01 g) of CO₂.

Next, we shall determine the number of mole of NaHCO₃ that will decompose to produce 25 g of CO₂. This can be obtained as follow:

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 44.01 g of CO₂.

Therefore, Xmol of NaHCO₃ will decompose to 25 g of CO₂ i.e

Xmol of NaHCO₃ = 25 / 44.01

Xmol of NaHCO₃ = 0.568 mole

1. Determination of the theoretical yield of NaOH.

From the balanced equation above,,

1 mole of NaHCO₃ decomposed to produce 40 g of NaOH.

Therefore, 0.568 mole of NaHCO₃ will decompose to produce = 0.568 × 40 = 22.72 g of NaOH.

Thus, the theoretical yield of NaOH is 22.72 g

2. Determination of the percentage yield of NaOH.

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Actual yield of NaOH = 5.03 g

Percentage yield of NaOH =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.03 / 22.72 × 100

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