<span>If you look at the chlorine box, with the symbol Cl, you see the atomic mass is equal to 35.453 atomic mass units. This is the weighted average mass of chlorine, including its isotopes, as found in nature. This also means that one mole of chlorine atoms has a mass of 35.453 grams.</span>
Answer:
4.5 g/L.
Explanation:
- To solve this problem, we must mention Henry's law.
- Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
- It can be expressed as: P = KS,
P is the partial pressure of the gas above the solution.
K is the Henry's law constant,
S is the solubility of the gas.
- At two different pressures, we have two different solubilities of the gas.
<em>∴ P₁S₂ = P₂S₁.</em>
P₁ = 525.0 kPa & S₁ = 10.5 g/L.
P₂ = 225.0 kPa & S₂ = ??? g/L.
∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.
0.953948 is what I got from dividing pressure value by 760.
According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.
The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.
Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases
\frac{a}{b}=\frac{15}{35}=\frac{10}{y}
rearranging,
y=\frac{10\times 35}{15}=23.3
Thus, mass of b produced will be 23.3 g.
Answer: Edge length of the unit cell = 628pm
Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is
Edge length of Unit cell (a) = (4R)/(√3)
R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m
a = (4 × (2.72 × (10^-10)))/(√3)
a = (6.28157 × (10^-10))m = 628pm
