For this problem, we use the formula for sensible heat which is written below:
Q= mCpΔT
where Q is the energy
Cp is the specific heat capacity
ΔT is the temperature difference
Q = (55.5 g)(<span>0.214 cal/g</span>·°C)(48.6°C- 23°C)
<em>Q = 304.05 cal</em>
Answer:
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Explanation:
In reduction-oxidation reaction two reactions take place, one is oxidation and the other is reduction reaction. In an oxidation reaction, there is the loss of an electron whereas in the reduction reaction there is gain of electron occus.
Reduction reaction occurs on the cathode, in a reduction of water there is gain of 2 electrons to gaseous hydrogen in basic aqueous solution. half-reaction for the reduction of liquid water to gaseous hydrogen in basic aqueous solution-
2H₂O (liq) + 2e⁻⇒ H₂ (g) + 2OH⁻ (aq)
Answer:
P2 = 19.2atm
Explanation:
Initial pressure (P1) = 16atm
Initial temperature (T1) = 340K
Final temperature (T2) = 408K
Final pressure (P2) = ?
This question involves the use of pressure law
Pressure law states that the pressure of a fixed mass of gas is directly proportional to it's temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
Therefore,
P1 / T1 = P2 / T2 = P3 / T3 = ......=Pn / Tn
P1 / T1 = P2 / T2
We need to solve for P2
P2 = (P1 × T2) / T1
Now we can plug in the values and solve for P2
P2 = (16 × 408) / 340
P2 = 6528 / 340
P2 = 19.2atm
The final pressure (P2) of the gas is 19.2atm
Where’s the problem I know how to do it
