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:
This question appear incomplete
Explanation:
This question appear incomplete. However, fuel is formed through a natural phenomenon involving the conversion of large amount dead and decayed organisms (usually algae and zooplanktons) to combustible fuel through exposure to relatively high temperature and pressure (over millions of years) in the earth's crust. Thus, since this involves a sort of absorption of heat energy (from the earth's crust), it can be referred to be an endothermic reaction.
An an increase in
temperature lead to more effective collisions between reactant particles and an
increase in the rate of a chemical reaction because the number of
molecules with sufficient energy to react increases. The answer is number 3.
Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
