Answer:
400cal
Explanation:
Since ice melted at 0 Celesius then the heat gained by ice (latent heat) will melt it so you should substitute in that law
Q=mlf ..where Q is the heat required to convert ice to water , m is the mass of ice and lf is the latent heat of fusion
Q=5∗80=400cal
Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
<h3>
<u>Answer;</u></h3>
= 930.23 mL
<h3><u>Explanation</u>;</h3>
Using the combined gas law;
P1V1/T1 = P2V2/T2
Where; P1 = 600 kPa, V1 = 800 mL, and T1 = -25 +273 = 258 K, and
V2= ?, P2 = 1000 kPa, and T2 = 227 +273 = 500 K
Thus;
V2 = P1V1T2/T1P2
= (600 ×800 ×500) / (258 × 1000)
= 930.23 mL
