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Vlad [161]
4 years ago
7

Calculate the percentage by mass of lead in pbco3.

Chemistry
2 answers:
Sonja [21]4 years ago
7 0

Answer:

The correct answer is d.

Explanation:

Step 1: First we must obtain the molecular weight of PbCO_{3} and Pb.

PM_{PbCO_{3} } = 267,21 \frac{g}{mol}

PM_{Pb} = 207,2 \frac{g}{mol}

Step 2: We can now calculate the percentage of lead in a PbCO_{3} molecule by stating a simple rule of three:

PM_{PbCO_{3} } -->    100%.

PM_{Pb}              -->    X

%Pb = \frac{PM_{Pb}}{ PM_{PbCO_{3} }} * 100%

%Pb = \frac{207.2 \frac{g}{mol}}{267.21 \frac{g}{mol}} * 100%

%Pb = 77,54 %

Have a nice day!

nignag [31]4 years ago
5 0
Pb/pbco3*100
207/207+12+48
207/267*100
=77.53%
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A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a
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<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = 62.364\text{ L. Torr }mol^{-1}K^{-1}

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0
3 years ago
How do the properties of compounds often compare with the properties of the elements that make them
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3 years ago
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Answer:

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3 years ago
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CaHeK987 [17]

Answer:

Q = 30355.2 J

Explanation:

Given data:

Mass of ice = 120 g

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Formula:

Q = m.c. ΔT

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m = mass of given substance

c = specific heat capacity of substance

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Q = 30355.2 J

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