Answer:
V₂ = 0.6 V.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n is constant, and have different values of P, V and T:
<em>(P₁V₁T₂) = (P₂V₂T₁).</em>
<em></em>
V₁ = V, P₁ = P, T₁ = T.
V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.
<em>∴ V₂ = (P₁V₁T₂)/(P₂T₁) =</em> (P)(V)(0.75 T)/(1.25 P)(T)<em> = 0.6 V.</em>
Answer:Water contains almost one thousand times more matter than the same volume of air would, therefore it is denser. ... If 500 mL of a liquid has a density of 1.11 g/mL, what is its mass? ... Density is the mass per unit of volume of a substance. ... A block of platinum with a mass of 4,290 kilograms has a volume of .2 cubic meters.
Explanation:
Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
<em>Change in heat:</em>
q = m*S*ΔT
<em>Where q is heat in J,</em>
<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
<em>Molar heat of solution:</em>
3830.87J/0.1048 moles KNO3 =
36554J/mol =
<h3>36.55kJ/mol</h3>
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Answer: Mass of silver deposited at the cathode is 37.1g
Explanation: According to Faraday Law of Electrolysis, the mass of substance deposited at the electrode (cathode or anode) is directly proportional to quantity of electricity passed through the electrolyte
Faraday has found that to liberate one gm eq. of substance from an electrolyte, 96500C of electricity is required.
+e− ==> Ag(s)
Given that
Current (I) = 8.5A
Time (t) = 65 *60 = 3900s
Quantity of electricity passed = 8.5*3900 =33150C
Molar mass of Ag= 108g
96500C will liberate 108g
33150C will liberate Xg
Xg= (108*33150)/96500
=37.1g
Therefore the mass of Ag deposited at the cathode is 37.1g.
