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Naily [24]
3 years ago
12

Prospectors are considering searching for gold on a plot of land that contains 2.45 g of gold per bucket of soil. If the volume

of the bucket is 5.25L, how many grams of gold are there likely to be in 1.0 cubic feet of soil
Chemistry
1 answer:
makvit [3.9K]3 years ago
8 0

Answer:

13.2 g of gold

Explanation:

We'll begin by converting 5.25 L to ft³.

This can be obtained as follow:

Recall:

1 L = 0.0353 ft³

Therefore,

5.25 L = 5.25 × 0.0353

5.25 L = 1.85×10¯¹ ft³

From the question given above,

2.45 g of gold is present in 5.25 L ( i.e 1.85×10¯¹ ft³) of soil.

Therefore, Xg of gold will be present in 1 ft³ of soil i.e

Xg of gold = 2.45/1.85×10¯¹

Xg of gold = 13.2 g

Therefore, 13.2 g of gold is present in 1 ft³ of the soil.

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when 1 mol of glucose is burned, 2802.5kj of energy is released. calculate rhe quantity of energy released to a person by eating
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77.78 kJ of energy is released when 1 mol of glucose is burned, 2802.5 kJ of energy is released.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 ×10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Calculate the moles of 5.00g of glucose.

Given mass = 5.00g

The molar mass of glucose = 180.156 g/mol

Moles = \frac{mass}{molar \;mass}

Moles = \frac{5.00g}{180.156 g/mol}

Moles =0.02775372455

The quantity of energy released to a person by eating 5.00g of glucose in a candy.

0.02775372455 x 2802.5 kJ

77.77981305 kJ =77.78 kJ

Hence, 777.78 kJ of energy is released.

Learn more about moles here:

brainly.com/question/8455949

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When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 14.4 g of carbon were burned in the presence of
Natasha2012 [34]

When carbon reacts with oxygen it forms CO2. This can depicted by the below equation.

C + O2→ CO2

It has been mentioned that when 14.4 g of C reacts with 53.9 g of O2, then 15.5 g of O2 remains unreacted. <u>This indicates that Carbon is the limiting reagent and hence the amount of CO2 produced is based on the amount of Carbon burnt.</u>

C + O2→ CO2

In the above equation , 1 mole of carbon reacts with 1 mole of O2 to produce 1 mole of CO2.

In this case 14.4 g of Carbon reacts with 53.9 of O2 to produce "x"g of CO2.

<u>No of moles = mass of the substance÷molar mass of the substance</u>

No of moles of carbon = 14.4 /12= 1.2 moles

No of moles of O2 = Mass of reacted O2/Molar mass of O2.

No of moles of O2 = (Total mass of O2 burned - Mass of unreacted O2)/32

No of moles of O2 = (53.9-15.5) ÷ 32 = 1.2 moles.

Hence as already discussed 1 mole of Carbon reacts with 1 mole of O2 to produce 1 mole of CO2. In this case 1.2 moles of carbon reacts with 1.2 moles of O2 to produce 1.2 moles of CO2.

Moles of carbon dioxide = Mass of CO2 produced /Molar mass of CO2

Mass of CO2 produced(x) = Moles of CO2 ×Molar mass of CO2

Mass of CO2 produced(x) = 1.2 x 44 = 52.8 g

<u>Thus 52.8 g of CO2 is produced.</u>

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