Plz help bc this s due at exactly 3:30pm
2 answers:
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5.732 grams of AgCl is formed when 0.200 L of 0.200 M AGNO3 reacts with an excess of CaCl2.
Explanation:
The balanced equation:
2 AgNO3(aq) + CaCl2(aq) -----> 2 AgCl(s) + Ca(NO3)2(aq)
data given:
volume of AgNO3 = 0.2 L
molarity of AgNO3 = 0.200 M
atomic weight of AgCl= 143.32 gram/mole
from the formula, number of moles can be calculated
Molarity =
number of moles of AgNO3 = 0.04
From the reaction:
2 moles of AgNO3 reacts to form 2 moles of AgCl
0.04 moles of AgNO3 reacts to form x mole of AgCl
=
= 0.04 moles of AgCl is formed
mass of AgCl formed is calculated by multiplying number of moles with atomic mass of AgCl
mass of AgCl = 0.04 x 143.32
= 5.732 grams of AgCl is formed.
Answer:
The answer to the question is
The temperature at which the vapor pressure will be 5.00 times higher than it was at 331 K is 353.0797 K.
Explanation:
To solve the question, we make use of the Clausius-Clapeyron equation as follows
Where P₁ = Initial pressure
P₂ = Final pressure
T₁ = Initial temperature = 331 K
T₂ = Final temperature
dvapH = ΔvapH = Heat of vaporization = 70.83 kJ / mol.
R = Universal gas constant = 8.3145. J K⁻¹ mol⁻¹
We are required to find the temperature when P₂ = 5 × P₁
Therefore we have
=
or T₂ =
= 353.0797 K
The vapor pressure be 5.00 times higher than it was at 331 K when the temperature is raised to 353.0797 K.