Answer:
0.29 mol/L
Explanation:
Its density is 1.029 g/ml so in a liter (1000 mL) there is 1029 g of solution, but only 5% is dextrose.
0.05x1029=51.45
So in a liter of D5W solution there is 51.45 g of dextrose.
Dextrose molar mass iss 180.156 g/mol, so in 51.45 g of dextrose there is
51.45/180.156=0.29 mol
In one liter of solution there is 0.29 mol of dextrose, so the molarity of such solution is 0.29 mol/L.
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
If two particles have the same charge they will repel each other when brought close together.
Answer:
The correct option is;
C. Old Age River
Explanation:
Among the three stages of the development of a river, which are the youthful, mature, and old age stages, the old age river is least dynamic
The water is very slow moving with a low gradient and lesser erosive power to alter the landscape which results in the appearance of flood planes
Examples of old age rivers include, lower Ganges, lower Nile, Indus, and Yellow rivers
Old age rivers are characterized by a broad shape, with a wide flood plane, a very gentle gradient and the water current is low.
Answer:
For the first oxide, 1 g gives 0.888 g of copper.
Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.
For the second oxide, 1 g gives 0.798 g of copper.
Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.
So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.
Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.
