These questions are all about indirect and direct variation with PV=nRT, the ideal gas equation
Q3.
false, because of PV=nRT, the ideal gas equation if V goes up, P has to go down to equal nRT
Q4. false, if V remains constant, and P and T are constant as moles of gas are added, then something is wrong becse something has to change when stuff is added (V has to go down)
Q5.
PV=nRT
when T and n are constant, (R is the gas constant)
PV=k, aka V=k/P which means inversly proportional
TRUE
Q6.
ggeasy
refer to past question
PV=k
if P is doubled then V has to halve in order to equal k
1/2 times 2=1
volume is halved
Q7. use charles law
V/T=k
so
given
V=4
T= kelvins, so 299
4/299=k
so when temp goes to 22 does V go to 3.95
4/299=3.95/295?
true
because they're equal
Q8
FALSE, must be used in kelvins
T=absolute tempurature in kelvins
Q9
PV=nRT
solve for T
(PV)/(nR)=T
use final volumes and pressures
P=5atm
V=24L
n=1
R=0.082057 atm L/(mol K)
(5atm*24L)/(1mol*0.082057 atm L/mol K)=T
see, if you didn't mess up, the units cancel nicely
T=1462.4
1200 K is closest
Q10
PV/T=constant because moles are constant (supposedly)
V=4L
P=2.08atm
T=275K
so find initial to final is constant
(2.08atm*4L)/(275K)=(Pfinal*2.5L)/(323K)
solve for Pfinal
Pfinal=3.92315 atm
answer is 3.9atm
Merry Christmas
Answer:
With an increase in temperature, the particles move faster as they gain kinetic energy, resulting in increased collision rates and an increased rate of diffusion. ... With an increase in temperature, the particles gain kinetic energy and vibrate faster and more strongly
Explanation:
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Answer:
See explanation
Explanation:
The acetoxy group in ortho position in phenyl acetate does interact with the phenyl moiety in the molecule via resonance.
This detailed interaction of the acetoxy group in ortho position in phenyl acetate with the phenyl moiety in the molecule via resonance is shown in the image attached.
This interaction is made possible because the oxygen atom of the acetoxy group has lone pairs of electrons that are suitably positioned to interact with the ring via resonance.
Answer:
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