5g of alumunium treated with excess of dilute HCl ,the volume of
hydrogen evolved at STP in dm3
Ans is 6.22......can u tell me how
euql masses of zinc n bromine were allowed to react till
completion of reaction to form ZnBr2 which substance is left
unreacted n to wat % of its original mass
ans-zinc-59.3 urgent plz,,,,,,,,,,,,,,,,
112al +6hcl = 2alcl3 +3 h2
Al. At.wt = 27
2* 27 = 3*2
5 = x
x =5*6/54
x = 5/9
x = 5/9
2g is 22.4l(dm3)
5/9 is 6.22 dm3
Therefore the answer is 6.22dm3
Editted after finding my mistake (stupid foolish mistake)
13572Al+6HCl→2AlCl3+3H2
mole of Al taken=5/27
moles of hydrogen formed=(5/27)*3/2=5/18
volume of hydrogen=(5/18)*22.4=6.22 dm3
112.
zn + br2 = znbr2
Therefore
65g of zinc will react with 160g of Br
Therefore we take 100g each
160 65
100 ?
100*65/160
x = 40.625g will be raeacted Therefor the remaining mass is 100- 40.625 = 59.375g
Therefore the percentage of the remainder is 59.375%
1under which condition the density of a gas is minimum
STP
0 degree C under 2 atm
100 degree c under 2 atm
100 degree c 0.5 atxplain
1357PV=nRT
P=(n/V)RT=(m/MV)RT=dRT/M
d=MP/RT
so d=KP/T
now try to find when it will be maximum
11D = M/V
DV is a constant ratio
Therefore if volume increases D reduces
Now
PV = nRT
nR is constant assume K
V = KT/P
Assume STP volume to be V0
Now in 2) Temperature is same as STP but pessure is increased therefore V1<V0
V 1 = K * 273 /2
3) V2>V1
V 2 = K * 373/2
Now check the no. 4.
V4 = K * 373 * 2
V0 = K * 273
Therefore V4>V0>V2>V1
Therefore The density will be minimum for 100°C and 0.5 atm
1on treatment with dilute alkali two molecules of acetone combine to form
acetone alcohol
diacetone
diactone alcohol
acetal
