Answer:
0.2024 M
Explanation:
For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
C 0 0 Initial
-2x +x +3x Reacts (stoichiometry is 1:1:3)
C - 2x x 3x Equilibrium
3x = 0.252
x = 0.084 M
The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.
Kc = ([H₂]³*[N₂])/([NH₃]²)
4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²
4.50 = 0.00533/(C - 0.168)²
4.50 = 0.00533/(C² - 0.336C + 0.028224)
4.50C² - 1.512C + 0.127008 = 0.00533
4.50C² - 1.512C + 0.121678 = 0
Solving the equation by a graphic calculator, for C > 0.168
C = 0.2024 M
This is the balanced eq
N2 + 3H2 -> 2NH3
first you need to find mole of N2 by using
mol = mass ÷ molar mass.
mol N2= 20g ÷ (14.01×2)g/mol
=0.7138mol
then look at the coefficient between H2 and NH3.
it is N2:NH3
1:2
0.7138:0.7138×2
0.7138:1.4276 moles
moles of NH3 = 1.4276 moles
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
Yes, if it’s a parallel circuit the wires are two different wires so it will light because that bulb isn’t connected to the one that went out
