It’s a polyatomic and it has a negative charge. It’s located in the nucleus of the atom, along with the protons.
Given:
Iron, 125 grams
T
1 = 23.5 degrees Celsius, T2 =
78 degrees Celsius.
Required:
Heat produced in kilojoules
Solution:
The molar mass of iron is 55.8
grams per mole. SO we need to change the given mass of iron into moles.
Number of moles of iron = 125 g/(55.8
g/mol) = 2.24 moles
<span>
Q (heat) = nRT = nR(T2 = T1)</span>
Q (heat) = 2.24 moles (8.314
Joules per mol degrees Celsius) (78.0 degrees Celsius – 23.5 degrees Celsius)
<u>Q (heat) = 1014.97 Joules or
1.015 kilojoules</u>
<span>This is the amount of heat
produced in warming 125 g f iron.</span>
<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
Answer:
Hello There!!
Explanation:
I believe the answer is c. Volume of the dilute solution is 100.00 mL.
hope this helps,have a great day!!
~Pinky~
