M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol
m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol
2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol
2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol
n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g
n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
<span>Seawater is water from a sea or ocean. On average, seawater in the world's oceans has a salinity of approximately 3.5% or 35 parts per thousand. This means that for every 1 liter (1000 mL) of seawater there are 35 grams of salts (mostly, but not entirely, sodium chloride) dissolved in it.
I used google
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Possibly, Yes if it is proven
Answer:
0.92 g
Explanation:
Given data:
Percentage of NaCl = 1.6%
Amount of NaCl present in 57.2 g = ?
Solution:
1.6% of 57.2 g NaCl:
1.6/100× 57.2 g
0.92 g
So in 57.2 g 1.6% of NaCl is 0.92 g.
N2+3H2---->2NH3
6.02×10 23 equals to one mole. if one mole of N2 enters reaction 2NH3 isproduced , so 2×6.02 10 23
12.04 × 10 23
