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2 years ago

How many oxygen atoms are there in 1665 grams of iron(II) phosphate? Please explain how you found your answer.

Chemistry

1 answer:

elena-14-01-66 [18.8K]2 years ago

8 0

Answer:

About 2.24 × 10²⁵ oxygen atoms

Explanation:

We want to determine the amount of oxygen atoms in 1665 grams of iron (II) phosphate (Fe₃(PO₄)₂).

To do so, we can convert from grams to moles of iron (II) phosphate, moles of iron (II) phosphate to moles of oxygen, and finally to atoms of oxygen.

Find the molar mass of iron (II) phosphate:

$\displaystyle \begin{aligned} \text{Molecular Weight} & = 3(55.85 + 2(30.97) + 8(16.00))\text{ g/mol} \\ \\ & = 357.49\text{ g/mol}\end{aligned}$

From the chemical formula, we can see that for each molecule of iron (II) phosphate, there are eight oxygen atoms. In other words, each mole of iron (II) phosphate has eight moles of oxygen atoms.

Finally, each mole of a substance contains 6.02 × 10²³ amount of that substances.

With the initial amount, multiply:

$\displaystyle \begin{aligned} 1665 \text{ g Fe$_3$(PO$_4$)$_2$} \cdot \frac{1 \text{ mol Fe$_3$(PO$_4$)$_2$}}{357.49 \text{ g Fe$_3$(PO$_4$)$_2$}}& \cdot \frac{8\text{ mol O}}{1 \text{ mol Fe$_3$(PO$_4$)$_2$}} \\ \\ &\cdot \frac{6.02\times 10^{23} \text{ atom O}}{1 \text{ mol O}} \\ \\ &=2.24\times 10^{25} \text{ atom O}\end{aligned}$

In conclusion, there are about 2.24 × 10²⁵ oxygen atoms in 1665 grams of iron (II) phosphate.

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Answer:

The proportion of dissolved substances in seawater is usually expressed in ppm, ppb or ppt

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The concentration of very diluted solutions should be expressed in parts per million, billion or trillion.

ppm = mass from the solute . 10⁶ / mass or volume of the solution

ppb = mass from the solute . 10⁹ / mass or volume of the solution

ppt = mass from the solute . 10¹² / mass or volume of the solution

ppm = mg/kg, μg/g, μg/mL → These are the units

ppb = ng/g

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3 years ago

If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?

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Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

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3 years ago

Activation process and importance of zymogen

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Answer:

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2 years ago

What is a Boitic factor

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Answer:

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3 years ago

You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of

Svetach [21]

Answer:

Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

Mass of Na₂SO₄ = 0.0773 g

Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

We place Na₂SO₄ = X and

K₂SO₄ = Y

Therefore

X +Y = 0.205 .........(1)

Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$

Amount of BaSO₄ from K₂SO₄ is;

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

$X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}$ = $X\times\frac{233.38 }{142.04}$ = 1.643·X

$Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}$ = $Y\times\frac{233.38 }{174.259 }$ = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

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