Answer:
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Explanation:
Mass of the solution = m
Mass of the KCN in solution = 779 mg
Mass by mass percentage of KCN solution = 0.173%
1 mg = 0.001 g
m = 450,289 mg × 0.001 g = 450.289 mg ≈ 450. g
450. g of 0.173 % KCN solution contains 779 mg of KCN.
Answer: Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.
Explanation:
According to Gay-Lussac's Law : 'The pressure of the gas increases with increase in temperature of the gas when volume of the gas is kept constant'.
At constant volume, pressure of the gas will decrease on decreasing the temperature or vice versa.
Decreasing the temperature inside the container will decrease the pressure of a gas inside a closed cubical container.
After 25 days, it remains radon 5.9x10^5 atoms.
Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.
N(Ra) = 5.7×10^7; initial number of radon atoms
t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days
n = 25 days / 3.8 days
n = 6.58; number of half-lifes of radon
N1(Ra) = N(Ra) x (1/2)^n
N1(Ra) = 5.7×10^7 x (1/2)^6.58
N1(Ra) = 5.9x10^5; number of radon atoms after 25 days
The half-life is independent of initial concentration (size of the sample).
More about half-life: brainly.com/question/1160651
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Answer:
Physical changes are changes that do not alter the identity of a substance. Chemical changes are changes that occur when one substance is turned into another substance.
Explanation:
The correct answer is option 2. A 0.8 M aqueous solution of NaCl has a higher boiling point and a lower freezing point than a 0.1 M aqueous solution of NaCl. This is explained by the colligative properties of solutions. For the two properties mentioned, the equation for the calculation of the depression and the elevation is expressed as: ΔT = -Km and <span>ΔT = Km, respectively. As we can see, concentration and the change in the property has a direct relationship.</span>
