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Dmitry [639]
3 years ago
8

A sample of gas at a pressure of 121.59 kPa, a volume of 31 L, and a temperature of 360 K contains how many moles of gas? A. 0.7

1 mol B. 0.96 mol C. 1.3 mol D. 1.8 mol
Chemistry
1 answer:
natali 33 [55]3 years ago
4 0

Answer:

C. 1.3 mol

Explanation:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is universal gas constant,

and T is absolute temperature.

Given:

P = 121.59 kPa

V = 31 L

T = 360 K

R = 8.3145 L kPa / mol / K

Find: n

n = PV / (RT)

n = (121.59 kPa × 31 L) / (8.3145 L kPa / mol / K × 360 K)

n = (3769.29 L kPa) / (2993.22 L kPa / mol)

n = 1.26 mol

Round to two significant figures, there are 1.3 moles of gas.

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Ah is gold and Rh is rhodium and Os is osmium and Ir iridium
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How many moles of copper are present in a sample that contains 4.55 x 10^24 atoms of copper
FinnZ [79.3K]

Hope you could understand.

If you have any query, feel free to ask.

3 0
2 years ago
One mole of acetyl chloride was mixed with one mole of dimethylamine. After the reaction is complete, what species can be found
fredd [130]

Answer:

N,N-dimethylacetamide is formed.

Explanation:

  • It is an example of a nucleophilic addition-elimination reaction. Here dimethylamine acts as a nucleophile.
  • In the first step, dimethyl amine gives nucleophilic addition reaction at carbonyl center of acetyl chloride.
  • In the second step, removal of Cl atoms occurs.
  • In the third step, deprotonation takes place from amino group to produce N,N-dimethylacetamide.
  • Full reaction mechanism has been shown below.

7 0
4 years ago
15.0 g of cream at 10.0 ℃ are added to an insulated cup containing 150.0 g of coffee at 78.6 °C. Calculate the equilibrium tempe
elena-14-01-66 [18.8K]

Answer:

The equilibrium temperature of the coffee is 72.4 °C

Explanation:

Step 1: Data given

Mass of cream = 15.0 grams

Temperature of the cream = 10.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 78.6 °C

C = respective specific heat of the substances( same as water) = 4.184 J/g°C

Step 2: Calculate the equilibrium temperature

m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)

15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)

62.76T2 - 627.6 = -627.6T2 + 49329.36

690.36T2 = 49956.96

T2 = 72.4 °C

The equilibrium temperature of the coffee is 72.4 °C

7 0
3 years ago
Linda performed the following trials in an experiment. Trial 1: Heat 30.0 grams of water at 0 °C to a final temperature of 40.0
nexus9112 [7]

<u>Answer:</u> The correct answer is Option b.

<u>Explanation:</u>

To calculate the amount of heat absorbed or released, we use the following equation:

q=mc\Delta T    .....(1)

where, q = amount of heat absorbed or released.

m = mass of the substance

c = heat capacity of  water = 4.186 J/g ° C      

\Delta T = Change in temperature

  • <u>For Trial 1:</u>

We are given:

m=30g\\\Delta T=[40-0]^oC=40^oC\\q=?J

Putting values in equation 1, we get:

q=30g\times 4.186J/g^oC\times 40^oC

q = 5023.2 J

  • <u>For trial 2:</u>

We are given:

m=40g\\\Delta T=[40-30]^oC=10^oC\\q=?J

Putting values in equation 1, we get:

q=40g\times 4.186J/g^oC\times 10^oC

q = 1674.4 J

Heat gained by Trial 1 than trial 2 = (5023.2-1674.4)J=3347J

Hence, the amount of heat gained in Trial 1 about 3347 J more than the heat released in Trial 2.

Thus, the correct answer is Option b.

4 0
3 years ago
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