Question

1. choose the best answer.

one way to shift equilibrium of a reaction is to change _____.
amount of time
concentration of reactant(s) or products
the reactants themselves
the amount of light exposure

1. balance the equation by typing the numbers into each blank. type in 1 if the coefficient is 1.

__ kclo₃ → kclo₄ + __ kcl

Explanation:

Question 6

To determine the correct answer, we analyze each option based on chemical equilibrium principles (Le Chatelier's principle). Changing the amount of time does not affect equilibrium (equilibrium is about concentrations, not time). Changing the reactants themselves isn't a standard way to shift equilibrium (we change their concentrations). Light exposure affects only photochemical reactions, not all equilibria. Changing the concentration of reactants or products is a key way (per Le Chatelier) to shift equilibrium.

Step1: Identify atoms to balance (K, Cl, O)

We have KClO₃ as reactant, KClO₄ and KCl as products. Let's denote coefficients as \( a \) (KClO₃), \( b \) (KClO₄), \( c \) (KCl).
For K: \( a = b + c \)
For Cl: \( a = b + c \) (same as K, since each has 1 Cl)
For O: \( 3a = 4b \)

Step2: Solve for coefficients

Let's find integers for \( a, b, c \). From O: \( 3a = 4b \), so least common multiple of 3 and 4 is 12. Let \( 3a = 12 \) ⇒ \( a = 4 \); \( 4b = 12 \) ⇒ \( b = 3 \). Then from K: \( 4 = 3 + c \) ⇒ \( c = 1 \).

Check:
Reactants: 4 K, 4 Cl, 12 O
Products: 3 K (KClO₄) + 1 K (KCl) = 4 K; 3 Cl (KClO₄) + 1 Cl (KCl) = 4 Cl; 3×4 O (KClO₄) = 12 O. Balanced.

Answer:

B. concentration of reactant(s) or products (assuming the options are labeled A, B, C, D with A: amount of time, B: concentration of reactant(s) or products, C: the reactants themselves, D: the amount of light exposure)

Question 7