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enot [183]
3 years ago
12

A sample of gas occupies a volume of 120.0 mL at a pressure of 0.75 atm and a temperature of 295 K. What will the volume be at a

pressure of 1.25 atm and a temperature of 345 K?
48 mL

4 mL

8 mL

84mL
Chemistry
1 answer:
MakcuM [25]3 years ago
6 0
The answer to the question is 84
(Step By Step Explanation).
P1 V1/T1= P2 V2/T2
::0.75 x 120/295 = 1.25 x V2/345
90/295= 1.25V2/345
(cross multiply )
1.25V2 x 295 = 90 x 345
368.75V2 = 31050
(divide by the coefficient of the unknown) .
368.75V2/368.75 = 31050/368.75
V2 = 84.2. = 84
::The volume is 82.
Mark As brainliest
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What is the mass of a sample of pure gold containing 3.00 x 1024 gold atoms?
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Answer: The mass is 980.6g of Gold.

Explanation:

We begin by looking for the number of moles equivalent to 3.0 x 10^24 gold atoms.

Using the Avogadro's number,

6.02 x 10^23 atoms of gold make up 1 mole of gold.

3.0 x 10^24 atoms would make up: 1 / 6.02 x 10^23 x 3.0 x 10^24 = 4.98moles.

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8 0
3 years ago
If a gold bar is 16,000 g mass and 800 cm^3 volume what is the density?
ki77a [65]

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<h3>The answer is 20 g/cm³</h3>

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density =  \frac{mass}{volume} \\

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5 0
4 years ago
A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr
sertanlavr [38]

Answer: The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.947 atm

P_2 = final pressure of gas = 0.987 atm

V_1 = initial volume of gas = 150 ml

V_2 = final volume of gas = 144 ml    

T_1 = initial temperature of gas = 25^0C=(25+273.15)K=298.15K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}

T_2=298.31K=(298.31-273.15)^0C=25.16^0C

The temperature of the gas at a pressure of  0.987 atm and volume of 144mL is 25.16^0C

4 0
3 years ago
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