Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C = 
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
![K = \frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)
where
K is given as ; 78.2 atm-1.
So, we have:
![78.2=\frac{[0.0530-x]}{[x][x]}](https://tex.z-dn.net/?f=78.2%3D%5Cfrac%7B%5B0.0530-x%5D%7D%7B%5Bx%5D%5Bx%5D%7D)
Using quadratic formula;
where; a = 78.2 ; b = 1 ; c= - 0.0530
= 
or 
= 
or 
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:
where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???
V = 1.64604
V ≅ 1.65 L
Answer:
Scientific Law
Explanation: Earth will ALWAYS attract objects to the center of itself because of gravity. This will never be disproven.
Answer:
3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.
Explanation:
The molecular mass of P2O5 is
P2 = 2* 31 = 62
O5 = 5 *<u> 16 = 80</u>
Molecular Mass = 142
Set up a Proportion
142 grams P2O5 supplies 62 grams of phosphorus
x kg P2O5 supplies 1.69 kg of phosphorus
Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.
142/x = 62/1.69 Cross multiply
142 * 1.69 = 62x combine the left
239.98 = 62x Divide by 62
239.98/62 = x
3.89 kg of P2O5 must be used.
Answer: The correct is 0.5 moles.
Explanation:
Answer:
1.047 M
Explanation:
The given reaction:
For dichromate :
Molarity = 0.254 M
Volume = 15.8 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 15.8 ×10⁻³ L
Thus, moles of dichromate :
Moles of dichromate = 0.0040132 moles
1 mole of dichromate react with 6 moles of iron(II) solution
Thus,
0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution
Moles of iron(II) solution = 0.02408 moles
Volume = 23 mL = 0.023 L
Considering:
<u>Molarity = 0.02408 / 0.023 = 1.047 M</u>
