If $\frac{Q}{K_{c}} < 1$ , then the reaction proceeds left to right but if $\frac{Q}{K_{c}} > 1$ , then the reaction proceeds, from right to left.
If true enter 1, if false enter 0.

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Solution

If $\frac{Q}{K_{c}} < 1$ , then the reaction proceeds left to right but if $\frac{Q}{K_{c}} > 1$ , then the reaction proceeds, from right to left.
When, $\frac{Q}{K_{c}} < 1$ , the value of the reaction quotient is less than the value of the equilibrium constant. Hence, the concentration of the products is lower than the equilibrium concentration. Hence, the equilibrium will shift in the forward direction.
When, $\frac{Q}{K_{c}} > 1$ , the value of the reaction quotient is more than the value of the equilibrium constant. Hence, the concentration of the products is more than the equilibrium concentration. Hence, the equilibrium will shift in the forward direction.

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