Answer: 116 g of copper
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 24.5A
t= time in seconds = 4.00 hr = 
(1hr=3600s)
of electricity deposits 63.5 g of copper.
352800 C of electricity deposits = 
of copper.
Thus 116 g of Cu(s) is electroplated by running 24.5A of current
Thus remaining in solution = (0.1-0.003)=0.097moles
Answer:
7.23 J
Explanation:
Step 1: Given data
- Mass of graphite (m): 566.0 mg
- Initial temperature: 5.2 °C
- Final temperature: 23.2 °C
- Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹
Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)
Q = 7.23 J
Answer:
Personally, I use Superglue
Explanation:
Answer:
from left to right: liquid, solid, gas
Explanation:
The answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---> Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
Hf C2H2 = 226.77
Hf Ca(OH)2 = -986.2
<span>Hf H2O = -285.83
Now,
</span><span>add them up. 226.77 - 986.2 + (2*285.83) = -187.77
</span><span>Add back the total enthalpy that is given in the question
-187.77+127.2 = -60.57 </span>
