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Kobotan [32]
3 years ago
13

The unheated Gas in the above system has a volume of 20.0 L at a temperature of 25.0 C and a pressure of 1.00 atm. The gas is he

ated to a temperature of 100.0 C while the volume remains constant. What is the pressure of the heated gas?
Chemistry
1 answer:
kipiarov [429]3 years ago
3 0

Answer:

1.25 atm.

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Initial volume (V1) = 20L

Initial temperature (T1) = 25°C

Initial pressure = 1 atm

Final temperature (T2) = 100°C

Final volume (V2) = constant i.e remain the same

Final pressure (P2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below:

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 25°C = 25°C + 273 = 298K

Final temperature (T2) = 100°C = 100°C + 273 = 373K

Step 3:

Determination of the final pressure of the gas. This is illustrated below:

Since the volume is constant, the following equation, P1/T1 = P2/T2 will be used to obtain the final pressure of gas as follow:

P1/T1 = P2/T2

Initial temperature (T1) = 298k

Initial pressure = 1 atm

Final temperature (T2) = 373K

Final pressure (P2) =?

P1/T1 = P2/T2

1/298 = P2 /373

Cross multiply to express in linear form

298 x P2 = 1 x 373

Divide both side by 298

P2 = 373/298

P2 = 1.25 atm.

Therefore, the pressure of the heated gas is 1.25 atm.

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Answer:

See explaination

Explanation:

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Please kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
A light bulb has a resistance of 96.8 Ω. What current flows through the bulb when it is connected to a 120 V source of electrica
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Answer:

I = 1.23 A

Explanation:

Given that,

The resistance of the lightbulb, R = 96.8 Ω

Voltage, V = 120 V

We need to find the current flows through the lightbulb. Let the current be I. We can use the ohm's law to find it i.e.

V=IR\\\\I=\dfrac{V}{R}\\\\I=\dfrac{120}{96.8}\\\\I=1.23\ A

So, the current flows through the bulb is 1.23 A.

8 0
3 years ago
What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

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3 years ago
Draw the structural formula for all the alkenes with the indicated molecular formula that, without undergoing a rearrangement, p
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Answer:

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Explanation:

while

C_{water}=4180j/kg*k

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3 0
3 years ago
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