Answer:
Explanation:
Solution,
For Johnathan
Distance(d)=800m
Time(t)=200sec
Now,
Speed=d/t=800/200=4m/s
For Abby
Distance(d)=200m
Time(t)=40sec
Now,
Speed=d/t=200/40=5m/s
Here, Abby is faster by 1m/s.
Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
Where,
is the Heat of vaoprization (J/mol)
is the normal boiling point of the gas (K)
is the Critical temperature of the gas (K)
is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:
Applying the above equation to find heat of vaporization as:
The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,
<u>Option C is correct</u>
Answer:
5. The mass of Na₂CO₃, that will produce 5 g of CO₂ is approximately 12.04 grams of Na₂CO₃
6. The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃ is 693. grams of N₂
Explanation:
5. The given equation for the formation of carbon dioxide (CO₂) from sodium bicarbonate (Na₂CO₃) is presented as follows;
(Na₂CO₃) + 2HCl → 2NaCl + CO₂ + H₂O
One mole (105.99 g) of Na₂CO₃ produces 1 mole (44.01 g) of CO₂
The mass, 'x' g of Na₂CO₃, that will produce 5 g of CO₂ is given by the law of definite proportions as follows;
The mass of Na₂CO₃, that will produce 5 g of CO₂, x ≈ 12.04 g
6. The chemical equation for the reaction is presented as follows;
N₂ + 3H₂ → 2NH₃
Therefore, one mole (28.01 g) of nitrogen gas, (N₂), reacts with three moles (3 × 2.02 g) of hydrogen gas (H₂) to produce 2 moles of ammonia (NH₃)
The mass 'x' grams of nitrogen gas (N₂) that will react completely with150 g of hydrogen (H₂) in the production of NH₃ is given as follows;
The mass of nitrogen gas (N₂) that will react completely with 150 g of hydrogen (H₂) in the production of NH₃, x = 693. grams
HCl = H⁺ + Cl⁻
c(HCl)=1.00*10⁻⁴ mol/l
[H⁺]=c(HCl)=1.00*10⁻⁴ mol/l
pH=-lg[H⁺]
pH=-lg(1.00*10⁻⁴)=4
pH=4
1. A - atoms
2. B - protons electrons and neutrons