The chemical equation given is:
<span>2x(g) ⇄ y(g)+z(s)</span>
Answer: the higher the amount of x(g) the more the forward reacton will occur and the higher the amounts of products y(g) and z(s) will be obtained at equilibrium.
Justification:
As Le Chatellier's priciple states, any change in a system in equilibrium will be compensated to restablish the equilibrium.
The higher the amount, and so the concentration, of X(g), the more the forward reaction will proceed to deal witht he high concentration of X(g), leading to an increase on the concentration of the products y(g) and z (s).
Use pv=nrT
where p is the pressure,
v is the volume,
n is the number of mole (which can be equal to mass /mr),
T is the temperature in kelvin,
and r is (molar constant) = 8.31 (units)
<span>The test dummy will continue forward until it makes contact with another object.</span>
0.001 would be the smallest.
Good Luck! :)
Answer:
65.08 g.
Explanation:
- For the reaction, the balanced equation is:
<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>
2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.
- Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:
<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>
<u><em>Using cross multiplication:</em></u>
2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.
0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.
∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.
<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol
) = <em>65.08 g.</em>
