Answer:
The pressure contribution from the heavy particles is 17.5 atm
Explanation:
According to Dalton's law of partial pressures, if there is a mixture of gases which do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture.
In the simulation:
the pressure of the 50 light particles alone was determined to be 5.9 atm, the pressure of the 150 heavy particles alone was measured to be 17.5 atm,
the total pressure of the mixture of 150 heavy and 50 light particles was measured to be 23.4 atm
Total pressure = partial pressure of Heavy particles + partial pressure of light particles
23.4 atm = partial pressure of Heavy particles + 5.9 atm
Partial pressure of Heavy particles = (23.4 - 5.9) atm
Partial pressure of Heavy particles = 17.5 atm
Therefore, the pressure contribution from the heavy particles is 17.5 atm
Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Answer:
B.Add acid to water,not water to acid
Explanation:
they should not be mixed
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
Answer:
[Na₂CO₃] = 0.094M
Explanation:
Based on the reaction:
HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)
It is possible to find pH using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [A⁻] / [HA]
Where [A⁻] is concentration of conjugate base, [CO₃²⁻] = [Na₂CO₃] and [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.
pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>
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Replacing these values:
10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]
<em> [Na₂CO₃] = 0.094M</em>
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