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EleoNora [17]
3 years ago
14

A gas has v= 4.0L, T =27 degree C and P=2.0 atm. What is the V if T=327 degree C and P=8.0 atm?

Chemistry
1 answer:
wariber [46]3 years ago
3 0
As I am reading the problem,i can tell the question gives you two temperatures, two pressures, one volume and asking for the other. this should be an indication that you need to use the following gas formula

P1V1/T1= P2V2/T2

P1= 2.0 atm
V2= 4.0 L
T1= 27= 300 K
P2= 8.0 atm
V2= ?
T2= 327= 600 k

let's plug in the values into the formula

(2.0 x 4.0)/ 300= (8.0 x V2)/ 600 K

V2= 2.0 Liters

answer is D
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A solution is made by mixing 55.g of thiophene C4H4S and 65.g of acetyl bromide CH3COBr.
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Answer:

Mole fraction of C₄H₄S = 0.55

Explanation:

Mole fraction is moles of solute / Total moles

Total moles are the sum of moles of solute + moles of solvent.

Let's find out the moles of our solute and our solvent.

Mass of solute: 55g

Mass of solvent: 65g

Mol = Mass / molar mass

55 g / 84.06 g/mol = 0.654 moles of C₄H₄S

65 g /123 g/mol = 0.529 moles of C₂H₃BrO

Total moles = 0.654 + 0.529 = 1.183 moles

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4 0
3 years ago
Which of the following is one part of a chemical formula?
melomori [17]

Answer:

C

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You would be correct in guessing C as in a chemical formula symbol of each element is stated.

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7 0
3 years ago
The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1
Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

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Where;

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t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

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t= 2.6/0.048

t= 54 days

3 0
3 years ago
What is atomic number of sodium​
FromTheMoon [43]

Answer:

Explanation:

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5 0
3 years ago
Read 2 more answers
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