Answer:
no idea with the answer pls check with otherr
Flammable liquid,gasoline, oil, and etc
Answer:
1.14 × 10³ mL
Explanation:
Step 1: Given data
- Initial volume of the gas (V₁): 656.0 mL
- Initial pressure of the gas (P₁): 0.884 atm
- Final volume of the gas (V₂): ?
- Final pressure of the gas (P₂): 0.510 atm
Step 2: Calculate the final volume of the gas
If we assume ideal behavior, we can calculate the final volume of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 0.884 atm × 656.0 mL/0.510 atm = 1.14 × 10³ mL
Answer:
333.7g of antifreeze
Explanation:
Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:
ΔT = Kf × m × i
Where:
ΔT is change in temperature (0°C - -20°C = 20°C)
Kf is freezing point depression constant (1.86°C / m)
m is molality of solution (moles solute / 0.5 kg solvent -500g water-)
i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)
Replacing:
20°C = 1.86°C / m × moles solute / 0.5 kg solvent × 1
5.376 = moles solute
As molar mass of ethylene glycol is 62.07g/mol:
5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.
