Answer:
Explanation:
according to the balance chemical equation
2 moles of c6h14 give us 14 moles of H2O
now mass(in gram) of 2 moles of C6H14 is
moles=mass/molar mass
mass=molar mass*moles
molar mass of C6H14 is 86 therefore
mass=86*2=172
now mass of 14 moles of water is
mass=molar mass*moles
molar mass of water is 18
mass=18*14=252
from the above calculations we conclude that
172 grams of C6H14 produce 252 grams of water
so 1.28 grams of C6H14 produce =252*1.28/172=1.87 grams of water
result is that 1.28 grams of C6H14 produced 1.87 grams of water
H2Se is the acid as it is donating H+ to H2O and forming SeH- (which is conjugate base).
H2O is the base since it’s receiving H+ from H2Se to form H3O+ (which is the conjugate acid)
Se is in the same group as O but is in different period which means Se has higher radius compred to O. this led to H-O bond being stronger than H-Se bond and H-Se bond which in turn makes it easier for H2Se to donate H to H2O
Solution :
We know, molecular mass of NH₃ is , M = 17 g/mol .
Number of moles of NH₃ in 7.2 × 10²⁴ are :
So, mass of NH₃ present in 11.96 moles is :
m = 11.96 × 17 gram
m = 203.32 gram
Hence, this is the required solution.
Answer:
c) B, C
Explanation:
NaOH(aq) + HBr(aq) -----> NaBr(aq) +H2O(l)
1) concentration of acid CA= 0.05 M
Concentration of base CB= 0.1 M
Volume of acid VA= 25.00ml
Volume of base VB= unknown
Number of moles of acid NA= 1
Number of moles of base NB= 1
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
VB= CAVANB/CB NB
VB= 0.05 × 25 × 1/ 0.1 ×1
VB= 12.5 ML
2.