Answer:
3.2 moles
Explanation:
First, we'll begin by writing a balanced equation for the Combustion of methane to produce carbon dioxide. This is illustrated below:
CH4 + 2O2 —> CO2 + 2H2O
From the balanced equation above,
1 mole of methane (CH4) reacted to produced 1 mole of carbon dioxide (CO2).
Therefore, 3.2 moles of methane (CH4) will react to produce 3.2 moles of carbon dioxide (CO2).
From the illustration above, 3.2 moles of methane is needed to produce 3.2 moles of carbon dioxide.
Al2(CrO4)3 = 401.9 grams per mole
Traditionally they include boron from group 3A, silicon and germanium in group 4A, aresnic and antimony in group 5A and tellurium from group 6A, although sometimes selenium, astatine, polonium and even bismuth have also been considered as metalloids. Typically metalloids are brittle and show a semi-metallic luster.
The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium. Five elements are less frequently so classified: carbon, aluminium, selenium, polonium, and astatine.
O valence electron number is the answer
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)
where
K is given as ; 78.2 atm-1.
So, we have:
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