Answer: 
Explanation:

Multiply by t on both sides.


Now divide by I to isolate t.


Answer:
energy released by a reaction is gained by the surroundings.
Explanation:
umm... I'm not 100% sure about this
1 mol of any substance is made of 6.022 x 10²³ units.
When its 1 mol of a formula, its made of 6.022 x 10²³ formula units
Therefore if 6.022 x 10²³ formula units of CeI₃ make up 1 mol of CeI₃
Then 9.42 x 10²¹ formula units make up - 1/(6.022 x 10²³) x (9.42 x 10²¹) = 0.0156 mol of CeI₃
Therefore number of moles of CeI₃ - 0.0156 mol
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT

Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then
)
Q= m*L

Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0
∘
C to 154
∘
C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT

Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
<u><em>The total heat required is 691,026.36 J</em></u>
You didn't put a + charge on one of the Ag. It has to be on the left side
2H2O + MnO2 ==> MnO4^- + 3e + 4H^+
Ag^+ + e ==> Ag^+
Those are the two BALANCED half reactions. Multiply equation 1 by 1 and equation 2 by 3 and add them.