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IceJOKER [234]
2 years ago
5

his is the chemical formula for chromium(III) nitrate: . Calculate the mass percent of oxygen in chromium(III) nitrate. Round yo

ur answer to the nearest percentage.
Chemistry
1 answer:
Alinara [238K]2 years ago
5 0

Answer:

\%\ Composition\ of\ iron=69.92\ \%

Explanation:

Percent composition is percentage by the mass of element present in the compound.

The formula for chromium(III) nitrate is Cr(NO_3)_3

Molar mass of chromium(III) nitrate = 238.011 g/mol

1 mole of chromium(III) nitrate contains 9 moles of oxygen

Molar mass of oxygen = 16 g/mol

So, Mass= Molar mass*Moles = 16*9 g = 144 g

\%\ Composition\ of\ iron=\frac{Mass_{iron}}{Total\ mass}\times 100

\%\ Composition\ of\ iron=\frac{144}{238.011}\times 100

\%\ Composition\ of\ iron=69.92\ \%

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Answer:

B. 45k

The human body is about 60 to 70% water.

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5 0
2 years ago
A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals an
MrRa [10]

Answer : 0.026 moles of oxygen are in the lung

Explanation :

We can solve the given question using ideal gas law.

The equation is given below.

PV = nRT

We have been given P = 21.1 kPa

Let us convert pressure from kPa to atm unit.

The conversion factor used here is 1 atm = 101.3 kPa.

21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm

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T = 295 K

R = 0.0821 L-atm/mol K

Let us rearrange the equation to solve for n.

n = \frac{PV}{RT}

n = \frac{0.208atm\times 3.0L}{0.0821 L.atm/mol K\times 295 K}

n = 0.026 mol

0.026 moles of oxygen are in the lung

3 0
2 years ago
Consider a saturated solution formed when 17.5 g of a solute dissolve in 28.3 g of a solvent, giving a total solution volume of
STALIN [3.7K]

Answer:

a) 38.2 % mass

b) 61.8 g solute/100 g solvent

c) 1.65 g/mL

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Given the data:

mass of solute = 17.5 g

mass of solvent= 28.3 g

total solution volume= 27.8 mL

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mass of solution = mass solute + mass solvent = 17.5 g + 28.3 g = 45.8 g

mass % = 17.5 g/45.8 g x 100 = 38.2 % mass

b)- solubility = grams of solute/ 100 g solvent

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7 0
2 years ago
A chemist is working on a reaction represented by this chemical equation:
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n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

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n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

6 0
2 years ago
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2 years ago
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