Answer:
NH₃
Explanation:
mass H = 6.10 grams
mass N = 28.00 grams
mass cpd = (6.10 + 28.00)grams = 34.10 grams
%H/100wt = (6.10/34.10)100% = 17.9% w/w
%N/100wt = (28.00/34.1)100% = 82.1% w/w
%/100wt => grams/100wt => moles => ratio => reduce => emp ratio
%H/100wt = 17.9% w/w => 17.9g => (17.9/1)moles = 17.9 moles H
%N/100wt = 82.1% w/w => 82.1g => (82.1/14)moles = 5.9 moles N
Ratio N:H => 17.9 : 5.9
Reduce mole ratio (divide by smaller mole value) => 17.9/5.9 : 5.9/5.9
=> 3HY:1H empirical ratio => empirical formula NH₃ (ammonia)
Answer:
The answer to your question is V = 19.9 L
Explanation:
Data
mass of Al = 8.60 g
volume of H₂ = ?
Balanced chemical reaction
2Al + 6HCl ⇒ 2AlCl₃ + 3H₂
1.- Use proportions to calculate the mass of H₂ produced
Atomic mass of Al = 2 x 27 = 54 g
Atomic mass of H₂ = 6 x 1 = 6 g
54 g of Al -------------------- 6 g of H₂
8 g of Al ------------------- x
x = (8 x 6) / 54
x = 48 / 54
x = 0.89 g of H₂
2.- Calculate the volume of H₂
- Convert the H₂ to moles
1 g of H₂ --------------- 1 mol
0.89 g ---------------- x
x = 0.89 moles
1 mol of H₂ -------------- 22.4 l
0.89 moles ------------- x
x = (0.89 x 22.4) / 1
x = 19.9 L
Answer:
i need a picture to solve
Explanation:
Answer:
5.004kg
Explanation:
Combustion of carbon
C+O2=CO2
from the relationship of molar ratio
mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)
mass of carbon =1000kg
atomic mass of carbon =12
volume of CO2 produced=1000×22.4/12
volume of CO2 produced =1866.6dm3
from the combustion reaction equation provided
CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)
applying the same relationship of molar ratio
no of mole of CO2=no of mole of urea
therefore
vol of CO2\22.4=mass of urea/molar mass of urea
molar mass of urea=60.06g/mol
from the first calculation
vol of CO2=1866.6dm3
mass of urea=1866.6×60.06/22.4
mass of urea=5004.82kg
Answer:
(upper right) corner of the periodic table to the bottom left corner
