Answer:
The standard cell potential of the reaction is 0.78 Volts.
Explanation:

Reduction at cathode :
Reduction potential of
to Cu=
Oxidation at anode:

Reduction potential of
to Fe=
To calculate the
of the reaction, we use the equation:

Putting values in above equation, we get:

The standard cell potential of the reaction is 0.78 Volts.
THE ANSWER
for solving this, we just use the formula of the equation of state
PV = n RT
where
P: pressure =10atm
V volume
n number of mole = 35.8 moles
R universal gas constant = 0.082
T: The temperature= 70°C= 343.15 Kelvin
therefore, for finding V
we have
V= (n RT) / P = 35.8 x 0.082 x 343.15 / 10 = 100.7 ≈ 101 L
the answer is V = 101L
<h3>
Answer:</h3>
89.88° C
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of gold cylinder as 75 g
- specific heat of gold is 0.129 J/g°C
- Initial temperature of gold cylinder is 65°C
- Mass of water is 500 g
- Initial temperature of water is 90 °C
We are required to calculate the final temperature;
- We know that Quantity of heat is given by the product of mass, specific heat capacity and change in temperature.
<h3>Step 1: Calculate the quantity of heat absorbed by the Gold cylinder</h3>
Assuming the final temperature is X° C
Then; ΔT = (X-65)°C
Therefore;
Q = 75 g × 0.129 J/g°C × (X-65)°C
= 9.675X - 628.875 Joules
<h3>Step 2: Calculate the quantity of heat released by water</h3>
Taking the final temperature as X° C
Change in temperature, ΔT = (90 - X)° C
Specific heat capacity of water is 4.184 J/g°C
Therefore;
Q = 500 g × 4.184 J/g°C × (90 - X)° C
= 188,280 -2092X joules
<h3>Step 3: Calculate the final temperature, X°C</h3>
we know that the heat gained by gold cylinder is equal to the heat released by water.
9.675X - 628.875 Joules = 188,280 -2092X joules
2101.675 X = 188908.875
X = 89.88° C
Thus, the final temperature is 89.88° C
Could you please mark me as brainliest?
Answer: 160.40 g Fe2O3 are needed.
Explanation:
Balanced equation: 2 Al + Fe2O3 —> Al2O3 + 2Fe
54.2 g Al * 1 mol Al / 26.98 g Al * 1 mol Fe2O3 / 2 mol Al * 159.69 g Fe2O3 / 1 mol = 160.40 g Fe2O3 are needed.
Heterogeneous mixture. cause you can separate the noodles from soup