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noname [10]
2 years ago
11

A student investigated the diffusion of ammonia gas, NH3, and hydrogen chloride gas, HCl. The damp red litmus paper for NH3 chan

ged colour after 30 seconds. How long does it take for the damp blue litmus paper to change colour for HCl?
Chemistry
1 answer:
Alex Ar [27]2 years ago
8 0

Answer:

43.96secs

Explanation:

M1 = molar mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

t1 = time for NH3 to diffuse = 30secs

M2 = molar mass of HCl = 1 + 35.5 = 36.5g/mol

t2 = time for HCl to diffuse =?

From Graham's law of diffusion:

t2/t1 = √(M2/M1)

t2/30 = √(36.5/17)

t2/30 = 1.465

Cross multiply

t2 = 30 x 1.465

t2 = 43.96secs

Therefore it will take 43.96secs for the damp blue litmus paper to change colour for HCl.

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A 150 W electric heater operates for 12.0 min to heat an ideal gas in a cylinder. During this time, the gas expands from 3.00 L
Brut [27]

<u>Answer:</u> The change in internal energy of the gas is 108.835 kJ

<u>Explanation:</u>

To calculate the work done for reversible expansion process, we use the equation:

W=P\Delta V=-P(V_2-V_1)

where,

W = work done

P = pressure = 1.03 atm

V_1 = initial volume = 3.00 L

V_2 = final volume = 11.0 L

Putting values in above equation, we get:

W=-(1.03)\times (11.0-3.00)=8.24L.atm=834.9J=0.835kJ     (Conversion factor:  1 L. atm = 101.325 J)

Calculating the heat from power:

Q=P\times t

where,

Q = heat required

P = power = 150 W

t =  time = 12 min = 720 s       (Conversion factor:  1 min = 60 s)

Putting values in above equation:

Q=150\times 720=108000J=108kJ

The equation for first law of thermodynamics follows:

Q=dU+W

where,

Q = total amount of heat required = 108 kJ

dU = Change in internal energy = ?

W = work done  = -0.835 kJ

Putting values in above equation, we get:

108kJ=dU+(-0.835)\\\\dU=(108+0.835)=108.835kJ

Hence, the change in internal energy of the gas is 108.835 kJ

7 0
2 years ago
What is the maximum number of grams of ammonia, nh3, which can be obtained from the reaction of 10.0 g of h2 and 80.0 g of n2? n
Lelechka [254]

<u>Answer:</u> The mass of ammonia produced is 28.22 g

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol

  • <u>For nitrogen gas:</u>

Given mass of nitrogen gas = 80.0 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen gas}=\frac{80.0g}{28g/mol}=2.86mol

The given chemical equation follows:

N_2+3H_2\rightarrow 2NH_3

By Stoichiometry of the reaction:

3 moles of hydrogen gas reacts with 1 mole of nitrogen gas

So, 5 moles of hydrogen gas will react with = \frac{1}{3}\times 5=1.66mol of nitrogen gas

As, given amount of nitrogen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of hydrogen gas produces 1 mole of ammonia

So, 5 moles of hydrogen gas will produce = \frac{1}{3}\times 5=1.66moles of ammonia

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 1.66 moles

Putting values in equation 1, we get:

1.66mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=(1.66mol\times 17g/mol)=28.22g

Hence, the mass of ammonia produced is 28.22 g

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2 years ago
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The wall would absorb extra heat during the day when the sun is out, then release the heat back into the room when the sun goes down.
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I’m pretty sure it’s d
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