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Nady [450]
3 years ago
15

How is an average mass different from a weighted mass

Chemistry
1 answer:
romanna [79]3 years ago
6 0

In an average mass, each entry has equal weight. In a weighted average, we multiply each entry by a number representing its relative importance.

Assume that your class consists of 15 girls and 5 boys. Each girl has a mass of 54 kg, and each boy has a mass of 62 kg.

<em>Average mass</em> = (girl + boy)/2 = (54 kg + 62 kg)/2 = <em>58 kg</em>

<em>Weighted average (Method 1) </em>

Use the <em>numbers of each</em> gender (15 girls + 5 boys) ,

Weighted average = (15×54 kg + 5×62 kg)/20 = (810 kg + 310 kg)/20

= 1120 kg/20 = <em>56 kg</em>.

If you put all the students on one giant balance, their total mass would be

1120 kg and the average mass of a student would be <em>56 kg. </em>

<em>Weighted average (Method 2) </em>

Use the <em>relative percentages</em> of each gender (75 % girls and 25 % boys).

Weighted average = 0.75×54 kg + 0.25×62 kg = 40.5 kg + 15.5 kg = <em>56 kg</em>

Each girl contributes 40.5 kg and each boy contributes 15.5 kg to the <em>weighted average</em> mass of a student.

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<u>Answer:</u>

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<u>For b:</u> The mixture will need to produce more reactants to reach equilibrium.

<u>For c:</u> The mixture will need to produce more products to reach equilibrium.

<u>Explanation</u>:

For the given chemical equation:

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The expression of K_{p} for above equation follows:

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We are given:

Value of K_p = 0.26

There are 3 conditions:

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For the given options:

  • <u>For a:</u>

We are given:

p_{NOCl}=0.11atm\\p_{NO}=0.16atm\\p_{Cl_2}=0.30atm

Putting values in expression 1, we get:

Q_p=\frac{(0.11)^2}{(0.16)^2\times 0.30}=1.57

As, K_{p}; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

  • <u>For b:</u>

We are given:

p_{NOCl}=0.048atm\\p_{NO}=0.12atm\\p_{Cl_2}=0.10atm

Putting values in expression 1, we get:

Q_p=\frac{(0.048)^2}{(0.12)^2\times 0.10}=1.6

As, K_{p}; the reaction is reactant favored

Hence, the mixture will need to produce more reactants to reach equilibrium.

  • <u>For c:</u>

We are given:

p_{NOCl}=5.20\times 10^{-3}atm\\p_{NO}=0.15atm\\p_{Cl_2}=0.15atm

Putting values in expression 1, we get:

Q_p=\frac{(5.20\times 10^{-3})^2}{(0.15)^2\times 0.15}=0.008

As, K_{p}>Q_p; the reaction is product favored.

Hence, the mixture will need to produce more products to reach equilibrium.

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