Answer:
to the left
Explanation:
<u>If the concentration of products is increased for a reaction that is in equilibrium, the equilibrium would shift to the left side of the reaction (the reactant's side). </u>
For a reaction that is in equilibrium, the reaction is balanced between the reactants and the products. According to Le Cha telier's principle, if one of the constraints capable of influencing the rate of reactions is applied to such a reaction that is in equilibrium, the equilibrium would shift so as to neutralize the effects created by the constraint.
<em>Hence, in this case, if the concentration of the products of a reaction in equilibrium is increased, the equilibrium would shift in such a way that more reactants are formed so as to annul the effects created by the increase in the concentration of the products. Since reactants are always on the left side of chemical equations, it thus means that the equilibrium would shift to the left.</em>
Answer:
mCO2= 49.6932 kgCO2
Explanation:
Hello! Let's solve this!
First we propose the balanced equation C3H8 + 5O2 ---> 3CO2 + 4H2O
We see that each mole of C3H8 (propane) we get 3 moles of CO2
From the propane volume we can obtain the grams of propane used.
molpropane = 26.5L * (1000mL / 1L) * (0.621g / 1mL) * (1mol / 44g) = 374.01mol propane
mCO2 = 374.01molC3H8 * (3molCO2 / 1molC3H8) * (44gCO2 / 1molCO2) = 49369.32g * (1kg / 1000g) = 49.6932 kgCO2
mCO2= 49.6932 kgCO2
Answer:
the water concentration at equilibrium is
⇒ [ H2O(g) ] = 0.0510 mol/L
Explanation:
- CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30
⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L
⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
replacing in Kc:
⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30
⇒ 0.0721 [ H2O(g) ] = 3.679 E-3
⇒ [ H2O(g) ] = 0.0510 mol/L
The masses of the objects and the distance between them
-hope it helps
Mass of KNO₃ : = 40.643 g
<h3>Further explanation</h3>
Given
28.5 g of K₃PO₄
Required
Mass of KNO₃
Solution
Reaction(Balanced equation) :
2K₃PO₄ + 3 Ca(NO₃)₂ = Ca₃(PO₄)₂ + 6 KNO₃
mol K₃PO₄(MW=212,27 g/mol) :
= mass : MW
= 28.5 : 212,27 g/mol
= 0.134
Mol ratio of K₃PO₄ : KNO₃ = 2 : 6, so mol KNO₃ :
= 6/2 x mol K₃PO₄
= 6/2 x 0.134
= 0.402
Mass of KNO₃ :
= mol x MW KNO₃
= 0.402 x 101,1032 g/mol
= 40.643 g
