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olchik [2.2K]
2 years ago
6

Assume that helium behaves as an ideal monatomic gas. If 2 moles of helium undergo a temperature increase of 100 k at constant p

ressure, how much work is done by the gas?
Chemistry
1 answer:
Helga [31]2 years ago
3 0

4157 J work is done by the gas , If 2 moles of helium undergo a temperature increase of 100 K at constant pressure.

<h3>What are Noble Gases ?</h3>

The noble gases are helium, argon, krypton, xenon, and radon, in order of their mass.

They are called noble gases because they are so majestic that they do not react with anything in general.

The work done can be calculated by the formula

Q=n Cp ΔT

Cp= ( 1+ f/2)R

R= 8.3144598 J. mol-1.

Cp = (1+3/2) * 8.314

Cp = 5*8.314/2

Q= 2 * 5 *8.314 *100/2

Q = 4157 J

Therefore 4157 J work is done by the gas , If 2 moles of helium undergo a temperature increase of 100 K at constant pressure.

To know more about Noble Gas

brainly.com/question/2094768

#SPJ4

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A 0.200 g sample of unknown metal x is dropped into hydrochloride acid and realeases 80.3 mL of hydrogen gas at STP using ideal
s344n2d4d5 [400]

Answer:

The number of mole of the unknown metal is 3.58×10¯³ mole

Explanation:

We'll begin by calculating the number of mole hydrogen gas, H2 that will occupy 80.3 mL at stp.

This is illustrated below:

Recall:

1 mole of any occupy 22.4L or 22400 mL at stp.

1 mole of H2 occupies 22400 mL at stp.

Therefore, Xmol of H2 will occupy 80.3 mL at stp i.e

Xmol of H2 = 80.3/22400

Xmol of H2 = 3.58×10¯³ mole

Therefore, 3.58×10¯³ mole of Hydrogen gas was released.

Now, we can determine the mole of the unknown metal as follow:

The balanced equation for the reaction is given below:

X + 2HCl —> XCl2 + H2

From the balanced equation above,

1 mole of the unknown metal reacted to produce 1 mole of H2.

Therefore, 3.58×10¯³ mole of the unknown metal will also react to produce 3.58×10¯³ mole of H2.

Therefore, the number of mole of the unknown compound is 3.58×10¯³ mole.

5 0
3 years ago
With the aid of a balanced chemical equation explain Chemical Reactions and the Types Of Chemical Reaction
e-lub [12.9K]

Answer:

Answers

1.)reactants: nitrogen and hydrogen; product: ammonia.

2.)reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water.

3.)N 2 + 3H 2 → 2NH 3

4.)Mg(OH) 2 + 2HNO 3 → Mg(NO 3) 2 + 2H 2O.

5.)2NaClO 3 → 2NaCl + 3O 2

6.)4Al + 3O 2 → 2Al 2O 3

7.)N 2(g) + 3H 2(g) → 2NH 3(g)

Explanation:

''.''

6 0
3 years ago
A sample of xenon gas occupies a volume of 6.80 L at 52.0°C and 1.05 atm. If it is desired to increase the volume of the gas sam
Pavlova-9 [17]

Answer:

207.03°C

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 6.80 L

T1 (initial temperature) = 52.0°C = 52 + 273 = 325K

P1 (initial pressure) = 1.05 atm

V2 (final volume) = 7.87 L

P2 (final pressure) = 1.34 atm

T2(final temperature) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:

P1V1/T1 = P2V2/T2

1.05 x 6.8/325 = 1.34 x 7.87/T2

Cross multiply to express in linear form as shown below:

1.05 x 6.8 x T2 = 325 x 1.34 x 7.87

Divide both side by 1.05 x 6.8

T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)

T2 = 480.03K

Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:

°C = K - 273

°C = 480.03 - 273

°C = 207.03°C

Therefore, the final temperature of the gas will be 207.03°C

5 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
Which is the most common nutrient deficiency in the world?
inna [77]
The most common nutrient deficiency in the world is Iron. 

Iron is the answer.

Final answer:- a) Iron
4 0
3 years ago
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