Answer:
20.1 g
Explanation:
The solubility indicates how much of the solute the solvent can dissolve. A solution is saturated when the solvent dissolved the maximum that it can do, so, if more solute is added, it will precipitate. The solubility varies with the temperature. Generally, it increases when the temperature increases.
So, if the solubility is 40.3 g/L, and the volume is 500 mL = 0.5 L, the mass of the solute is:
40.3 g/L = m/V
40.3 g/L = m/0.5L
m = 40.3 g/L * 0.5L
m = 20.1 g
Answer:
It makes the pasta to get hot faster and boil quicker.
Explanation:
Adding salt to water actually raises the boiling point of the water, due to a phenomenon called boiling point elevation. Essentially, adding any non-volatile solute such as salt to a liquid causes a decrease in the liquid’s vapour pressure. A liquid boils when the vapour pressure above it equals atmospheric pressure, so a lower vapour pressure means you need a higher temperature to boil the water. The reason salt makes water boil faster has to do with specific heat capacities, or the energy it takes to raise the temperature of a substance. Salt ions dissolved in water bind to water molecules, holding them stable and making it harder for them to move around. As a result, the non-salt bound water molecules receive more of the energy provided by the stove, and therefore they get hot faster and boil quicker.
Hi there!
Rutherford's experiment showed that the atom's nucleus <em>did </em>have a positive charge. The α particles' trajectory was altered when it went through the gold foil, which, in turn resulted in the discovery of the atom's positive charge.
I hope this helps!
Brady
Answer:
%Ionization = 1.63%
Explanation:
Hydrazine in aqueous media theoretically forms a difunctional hydroxyl system. However, for this problem assume only monofunctional ionization occurs. A second hydroxyl ionization would not likely occur as the formal cationic charge formed in the 1st ionization would inhibit a second ionization.
H₂NNH₂ + 2H₂O => HONHNHOH => HONHNH⁺ + OH⁻; Kb = 1.3 x 10⁻⁶
So, assuming all OH⁻ and HONHNH⁺ are delivered in the 1st ionization then a good estimate of the %ionization can be calculated.
HONHNHOH => HONHNH⁺ + OH⁻
C(i) => 0.490M 0M 0M
ΔC => -x +x +x
C(eq) => 0.490 - x x x
≅0.490M* => *x is dropped as Conc H₂NNH₂/Kb > 100
Kb = [HONHNH⁺][OH⁻]/[HONHNHOH]
1.3 x 10⁻⁶ = x²/0.490
=> x = [OH⁻] = [HONHNH⁺] = √[(1.3 x 10⁻⁶)(0.490)] = 8 x 10⁻⁴
=> %Ionization = (x/0.490)100% = (8 x 10⁻⁴/0.490)100% = 1.63%
