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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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Assume that a daily diet of 2000 calories (i.e. 8.37 x 106 J) is converted completely to body heat.
slava [35]

Answer:

(a) the mass of the water is 3704 g

(b) the mass of the water is 199, 285.7 g

Explanation:

Given;

Quantity of heat, H= 8.37 x 10⁶ J

Part (a) mass of water (as sweat) need to evaporate to cool that person off

Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg

H = m x Lvap.

m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg

mass in gram ⇒ 3.704 kg x 1000g = 3704 g

Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J

specific heat capacity of water, C, 4200 J/kg.°C

H = mcΔθ

where;

Δθ is the change in temperature = 35 - 25 = 10°C

m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

5 0
3 years ago
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sergeinik [125]

Answer:

Bohr model A

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A gas sample has a temperature of 22c with an unknown volume. The same gas has a volume of 456 mL when the temperature is 86c wi
Mkey [24]

The initial volume is 116.65 mL

<u>Explanation:</u>

<u />

Given:

Temperature, T₁ = 22°C

T₂ = 86°C

Volume, V₂ = 456 m

V₁ = ?

According to Charle's law:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

Substituting the values:

\frac{V_1}{22} =\frac{456}{86} \\\\V_1 = \frac{456 X 22}{86} \\\\V_1 = 116.65 mL

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Which of the following alcohols will give a positive chromic acid test?
Sindrei [870]
The answer is both B and C

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cyclohexanol can be oxidized to become cyclohexanone

and pentan-3-ol can be oxidized to become pentan-3-one

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