moles of hydrogen in reaction =
=
= 0.996 mol
ratio of
: Na is 1 : 2
∴ if mol of
= 0.996 mol
then mol of Na = (0.996 mol * 2)
= 1.99 mol
Mass of Na = molar mass * mol
= (22.99 g / mol) * (1.99 mol)
= 45.78 g
Note:
1) Molar Mass is the mass of an element measured in grams.
2) According to Avogadro's Law, the same volume of different gases at the same condition of Temperature and Pressure contain the same number of particles. From this law we know that at STP, one mole of gas occupies 22.4 L of volume (molar volume).
3) By calculating the mole of one specie in a reaction, one can use mole ratio based on the stoichimectric values (values used to balance the equation) given to the species upon balancing the equation to find the moles of another species.
Answer:
Molar mass of NH3,
MM = 1*MM(N) + 3*MM(H)
= 1*14.01 + 3*1.008
= 17.034 g/mol
mass(NH3)= 25.0 g
use:
number of mol of NH3,
n = mass of NH3/molar mass of NH3
=(25 g)/(17.03 g/mol)
= 1.468 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 35.8 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(35.8 g)/(32 g/mol)
= 1.119 mol
The balanced chemical equation is:
4 NH3 + 5 O2 ---> 4 NO + 6 H2O
4 mol of NH3 reacts with 5 mol of O2 for 1.468 mol of NH3, 1.835 mol of O2 is required
But we have 1.119 mol of O2
so, O2 is limiting reagent
we will use O2 in further calculation
Molar mass of NO,
MM = 1*MM(N) + 1*MM(O)
= 1*14.01 + 1*16.0
= 30.01 g/mol
According to balanced equation
mol of NO formed = (4/5)* moles of O2
= (4/5)*1.119 = 0.895 mol
use: mass of NO = number of mol * molar mass
= 0.895*30.01
= 26.86 g
Answer: 26.9 g
Answer:81.6%
Explanation:
Mass of CaCO3=4.010 g
Molar mass of CaCO3= 40+12+(16×3) = 100 g/mol.
Recall: number of moles(n)= mass÷ molar mass.
n=4.010÷100 = 0.0401 mol.
Molar mass of CaCl2 = 40+71= 111 g/mol.
Number of mol of CaCl2 = 5.455÷111= 0.04914 g/mol.
Mass of CaCl2 = 0.0401 × 111 = 4.4511 g of CaCl2.
Percent by mass of CaCl2 = (4.4511÷5.455) × 100
= 0.815967 ×100 = 81.5967%
Approximately; 81.6%.
Answer:
The molar heat of vaporization of dichloromethane is 30.8kJ/mole
Explanation:
Using Clausius Clapeyron equation
ln (P1/P2) = (ΔHvap/R) (1/T2-1/T1)
At initial temperature of Ooc , the vapour pressure is 134mmHg
Therefore T1 = 0+273 =273K
And P1 = 134mmHg
At normal boiling point of 40oC , the vapour pressure is 760mmhg
T2 = 40 +273 = 313K
P2 = 760mmHg
ln (134/760) = ΔHvap/(8.3145 J/molK)
( 1/313K - 1/273K)
ΔHvap = 30800J/mol
= 30.8kJ/mol
Therefore, the molar heat of vaporization can be calculated using
Clausius Clapeyron equation using the above steps
Not sure but i think living things adapt to their environment.
Adaptations are traits giving an organism an advantage in a certain environment. And variations of individuals is important for a healthy species. So i think adaptation might be the right answer because animal will try to adapt the certain environment. I hope this answer will help you.