Question

1. A 25.0-mL sample of chlorine gas (Cl2), with a temperature of 15 °C, is warmed up to 35 °C. What is its new volume?
2. 10.0 L of oxygen (O2) is subjected to a change in pressure, going from 0.75 atm to 2.35 atm. How is the volume affected?
3. The pressure of a gas sample decreases from 775 mm Hg (1 atm = 760 mm Hg) to 745 mm Hg. If its temperature before the change was 30. °C, what is it now?

Answer 1

any idea ?? which equation would you use for first one ??

Answer 2

\[\frac{ V1 }{ T1}=\frac{ V2 }{ T2}\]

Answer 3

@Nnesha

Answer 4

yes that's correct. Charles' law
plugin given values solve for v_2

Answer 5

don't forget to change temperature to Kelvin by adding 273 into C
I would also convert volume ml to L

Answer 6

let me know what you get or if you have any question feel free to ask :=))

Answer 7

V1:25.0
T1:15+273= 288 K
T2: 35+273=308 K
@Nnesha

Answer 8

good

Answer 9

\[\large\rm \frac{ V_1 }{ T_1 }=\frac{V_2}{V_2}\]
plugin numbers

Answer 10

there should be \[\rm T_2\] 2nd fraction

Answer 11

\[\frac{ 25.0 }{ 288 }=\frac{ V2 }{ 308 }\]

Answer 12

\[V2=\frac{ 25.0*308 }{ 288 }=26.74 L\]

Answer 13

@Nnesha

Answer 14

hmm it is good to write the unit as well so you won't make any mistake \[\huge\rm V_2 =\frac{25.0 mL \cdot 308 k}{288 k}\]
given volume is in mL

Answer 15

first of all you can cancel out the `K` \[\large\rm v_2=\frac{ 25.0 m L \ \cdot 308 \cancel{ k} }{ 288 \cancel{k }}\]
\[\large\rm v_2=\frac{ 25.0 m L \ \cdot 308 }{ 288 }\]now multiply 25.0 times 308 = 7700 \[\large\rm v_2=\frac{ 25.0 m L \ \cdot 308 }{ 288 } =\frac{7700 mL}{288}\]
simplify the fraction first and then convert mL to L

Answer 16

26.74 L

Answer 17

what is 7700/288 ??

Answer 18

26.73

Answer 19

right 7700/288 =26.74 mL not L
you still have to convert mL to L
1ml= .001 L

Answer 20

since we need volume in liter
therefore write litter at the top \[\huge\rm 26.7 ml ~(\frac{.001L}{1 mL})\]

Answer 21

0.0267

Answer 22

0.0267 L don't forget the unit