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Anni [7]
3 years ago
6

The density of lead is 11.3g/cm^3 and the density of chromium is 7.14g/cm^3. If we have one gram of each metal, what is the rati

o of the volume of lead to the chromium?
Chemistry
1 answer:
Lady bird [3.3K]3 years ago
5 0

The ratio of the volume of lead to the volume of chromium is 22 : 35

Density is simply defined as the mass of a substance per unit volume of the substance.

<h3>Density = mass / volume </h3>

To solve the question given above, we'll begin by calculating the volume of lead and chromium. This can be obtained as follow:

<h3>For Lead:</h3>

Mass of lead = 1 g

Density of lead = 11.3 g/cm³

<h3>Volume of lead =? </h3>

Density = mass /volume

11.3 = 1 / Volume

Cross multiply

11.3 × Volume = 1

Divide both side by 11.3

Volume = 1 / 11.3

<h3>Volume of lead = 0.088 cm³</h3>

<h3>For Chromium:</h3>

Mass of chromium = 1 g

Density of chromium = 7.14 g/cm³

<h3>Volume of chromium =? </h3>

Density = mass /volume

7.14 = 1 / Volume

Cross multiply

7.14 × Volume = 1

Divide both side by 7.14

Volume = 1 / 7.14

<h3>Volume of chromium = 0.14 cm³</h3>

Finally, we shall determine the ratio of the volume of lead to the volume chromium. This can be obtained as follow:

Volume of lead = 0.088 cm³

Volume of chromium = 0.14 cm³

<h3>Ratio of the volume of lead to chromium =? </h3>

Ratio = Volume of lead / Volume of chromium

Ratio = 0.088 / 0.14

Ratio = 22 / 35

<h3>Ratio = 22 : 35</h3>

Therefore, the ratio of the volume of lead to the volume chromium is 22 : 35

Learn more: brainly.com/question/17262276

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Now, I am a medical student and we have never had to convert a BP (blood pressure) to atm from mmHg, only ever kPA. SO, I am going to take a guess here and say that when you do the work to solve this, you are going to convert the Systolic (upper #) which is the 145. You should get 0.190789 and then convert the Diastolic (lower #) which is 65. You should get 0.08552632.

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(Just to note that is way to low of a BP, although it is irrelevant) Best wishes and good luck. "Remember, never just look for the right answer, look for why it is the right answer!"

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