Answer:
0.164541341 g H2
Explanation:
1) Convert grams to moles by dividing by RMM of Magnesium (24.31g).
2g Mg * (1 mol Mg / 24.31 g Mg) = 0.082270671 mol of Mg
2) Use the balanced equation's ratio of 1 mol Mg: 1 mol H2.
0.082270671 mol of Mg = 0.082270671 mol of H2
3) Convert the mol of H2 back into grams by multiplying by H2's RMM (2 g).
0.082270671 mol of H2 * 2 g H2 = 0.164541341 g H2
* Answer can be rounded to your liking *
The freezing point depression is calculated through the equation,
ΔT = (kf) x m
where ΔT is the difference in temperature, kf is the freezing point depression constant (1.86°C/m), and m is the molality. Substituting the known values,
5.88 = (1.86)(m)
m is equal to 3.16m
Recall that molality is calculated through the equation,
molality = number of mols / kg of solvent
number of mols = (3.16)(1.25) = 3.95 moles
Then, we multiply the calculated amount in moles with the molar mass of ethylene glycol and the answer would be 244.9 g.
I'm guessing it's C-Oxygen because in order to burn something, or create fire, oxygen HAS to be present.
The intermolecular force that attracts two nonpolar molecules is London dispersion forces, which are also called induced dipole-induced
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
