Answer:81.6%
Explanation:
Mass of CaCO3=4.010 g
Molar mass of CaCO3= 40+12+(16×3) = 100 g/mol.
Recall: number of moles(n)= mass÷ molar mass.
n=4.010÷100 = 0.0401 mol.
Molar mass of CaCl2 = 40+71= 111 g/mol.
Number of mol of CaCl2 = 5.455÷111= 0.04914 g/mol.
Mass of CaCl2 = 0.0401 × 111 = 4.4511 g of CaCl2.
Percent by mass of CaCl2 = (4.4511÷5.455) × 100
= 0.815967 ×100 = 81.5967%
Approximately; 81.6%.
Because if no one gets sick the hospitals will get no patients to cure and there would be overpopulation.
P Ar = 1.5
P Xe = 0.5
P tot = 2 atm
<h3>Further explanation </h3>
Dalton's law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases
Can be formulated:
P tot = P1 + P2 + P3 ....
The partial pressure is the pressure of each gas in a mixture
3 L Argon , P = 1 atm and 1 L Xe, P=1 atm⇒assume T = 273 K
mol Argon :
mol Xenon :
P tot = 1.5 + 0.5 = 2 atm
Answer:
1.85 g
Explanation:
The strategy here is to utilize the Henderson-Hasselbach equation
pH = pKa + log [A⁻] / [HA]
to calculate the ratio log [A⁻] / [HA], and from there to calculate the concentration [A⁻] and finally the mass of NaNO₂ from the number of moles assuming the final buffer volume is 50.0 mL ( that is the volume does not change by the addition of NaNO₂)
pH = pKa + log [NO₂⁻]/[HNO₂]
3.13 = 3.40 + log [NO₂⁻]/[HNO₂]
- 0.27 = log [NO₂⁻]/[HNO₂]
taking the inverse log function to both sides of this equation
0.54 = [NO₂⁻]/[HNO₂]
Now [HNO₂] = 1.0 M, therefore [NO₂⁻] = [NaNO₂] =
0.54 x 1.0 M = 0.54 M
from M = mol / L we get
mol = 0.54 mol/L x 0.050L = 0.027 mol
the molar mass of NaNO₂ is = 68.99 g / mol, so the mass of 0.027 mol is
0.027 mol x 68.99 g/mol = 1.85 g