Answer: 5.66 dm3
Explanation:
Given that:
Volume of neon gas = ?
Temperature T = 35°C
Convert Celsius to Kelvin
(35°C + 273 = 308K)
Pressure P = 0.37 atm
Number of moles N = 0.83 moles
Note that Molar gas constant R is a constant with a value of 0.0082 ATM dm3 K-1 mol-1
Then, apply ideal gas equation
pV = nRT
0.37atm x V = 0.83 moles x 0.0082 atm dm3 K-1 mol-1 x 308K
0.37 atm x V = 2.096 atm dm3
V = (2.096 atm dm3 / 0.37atm)
V = 5.66 dm3
Thus, the volume of the neon gas is 5.66 dm3
The mass of a sample of alcohol is found to be = m = 367 g
Hence, it is found out that by raising the temperature of the given product, the mass of alcohol would be 367 g.
Explanation:
The Energy of the sample given is q = 4780
We are required to find the mass of alcohol m = ?
Given that,
The specific heat given is represented by = c = 2.4 J/gC
The temperature given is ΔT = 5.43° C
The mass of sample of alcohol can be found as follows,
The formula is c = 
We can drive value of m bu shifting m on the left hand side,
m = 
mass of alcohol (m) = 
m = 367 g
Therefore, The mass of the given sample of alcohol is
m = 367g
It requires 4780 J of heat to raise the temperature by 5.43 C in the process which yields a mass of 367 g of alcohol.
- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
Answer:
g NaCl = 424.623 g
Explanation:
<em>C</em> NaCl = 3.140 m = 3.140 mol NaCl / Kg solvent
∴ solvent: H2O
∴ mass H2O = 2.314 Kg
mol NaCl:
⇒ mol NaCl = (3.140 mol NaCl/Kg H2O)×(2.314 Kg H2O) = 7.266 mol NaCl
∴ mm NaCl = 58.44 g/mol
⇒ g NaCl = (7.266 mol NaCl)×(58.44 g/mol) = 424.623 g NaCl
