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mr Goodwill [35]
1 year ago
11

What is the weighted average of a nail in the sample data given?

Chemistry
1 answer:
Natalka [10]1 year ago
5 0

The weighted average of the nail in accordance with the given data is 11.176g.

<h3>How to calculate weighted average?</h3>

Weighted average is an arithmetic mean of values biased according to agreed weightings.

The weighted average of the nail in the image above can be calculated by multiplying the decimal abundance with the mass of the nail, then summed up as follows;

Weighted average = (decimal abundance × mass 1) + (decimal abundance × mass 2)

Weighted average = (0.12 × 3.3) + (0.88 × 12.25)

Weighted average = 0.396 + 10.78

Weighted average = 11.176g

Therefore, 11.176g is the weighted average of the nail

Learn more about weighted average at: brainly.com/question/28042295

#SPJ1

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<span>LiOCH3</span>

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A pH of 15.0<br> a. is not possible.<br> b. is a base.<br> c. is an acid.<br> d. is a neutral.
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A gas with a volume of 8 L at a temperature of 473K is switched to a new container with a temperature of 350K. If the pressure r
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Answer:

The new  volume is 5.92 L,  which is approximately 6 L

Explanation:

As the gas temperature increases, the molecules move faster and take less time to reach the walls of the container. This means that the number of crashes per unit of time will be greater. That is, there will be an increase (for an instant) in the pressure inside the container and the volume will increase.

So Charles's Law is one of the gas laws that relates the volume and temperature of a certain amount of gas at constant pressure and says that:

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  • If the temperature decreases the volume decreases

Mathematically this is expressed by:

\frac{V}{T}=k

When you want to study two different states, an initial and a final one of a gas, this law is expressed by:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1: 8 L
  • T1: 473 K
  • V2: ?
  • T2: 350 K

Replacing:

\frac{8 L}{473K} =\frac{V2}{350K}

Solving:

V2=350K*\frac{8L}{473K}

V2=5.92 L

<u><em>The new  volume is 5.92 L,  which is approximately 6 L</em></u>

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