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What would happen to the rate of a reaction with rate law rate k NO H2 if the concentrations of NO were doubled?
The rate would be four times larger
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What would happen to the rate of a reaction with rate law rate k (NO2) (H2) if the concentration of Anna were doubled?
The rate of the reaction with the rate law ( \text{rate} = k[\text{NO}_2][\text{H}_2] ) is dependent solely on the concentrations of NO2 and H2. If the concentration of Anna, which is not included in the rate law, is doubled, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged unless the rate law itself involves Anna in some way, which is not indicated here.
What would happen to the rate of a reaction with rate law rate kno2h3 if the concentration of no were doubled?
If the concentration of NO was doubled in the rate law rate = k[NO]2[H3], the rate of the reaction would increase by a factor of 4. This is because the rate of a reaction typically increases with an increase in the concentration of reactants, raised to a power dictated by their respective coefficients in the rate law equation.
What would happen to the rate of a reaction with rate law rate k NO2H2 if NO were halved?
In the rate law given as rate = k[NO2][H2], the concentration of NO does not appear, so the rate of the reaction is independent of its concentration. Therefore, if the concentration of NO were halved, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged as long as the concentrations of NO2 and H2 remain constant.
If the volume of water produced during the reaction doubled what would happen to the ratio of hydrogen and oxygen?
If the volume of water produced during the reaction doubled, it would indicate that the reaction produced twice as much water as before. Since water (H₂O) is composed of two hydrogen atoms and one oxygen atom, the ratio of hydrogen to oxygen in the water remains constant at 2:1. Therefore, even though the volume of water increases, the ratio of hydrogen to oxygen in the reaction does not change; it remains the same.
Which equation would be used to calculate the rate constant from inital concentrations?
To calculate the rate constant (k) from initial concentrations, you would typically use the rate law equation for the reaction, which is expressed as ( \text{Rate} = k[A]^m[B]^n ), where ( [A] ) and ( [B] ) are the initial concentrations of the reactants, and ( m ) and ( n ) are their respective reaction orders. By measuring the initial rate of the reaction and substituting the initial concentrations into the rate law, you can rearrange the equation to solve for the rate constant ( k ).
What would happen to a System at equilibrium if more of one compound in a reaction were added according to le chateliers principle?
All concentrations would change (apex)
What would happen to the rate of a reaction with rate law rate kno2h3 if the concentration of no were doubled?
If the concentration of NO was doubled in the rate law rate = k[NO]2[H3], the rate of the reaction would increase by a factor of 4. This is because the rate of a reaction typically increases with an increase in the concentration of reactants, raised to a power dictated by their respective coefficients in the rate law equation.
What happens when sodium reacts wth sodium hydroxide?
nothing would happen. As there is sodium in both compounds nothing would react. At most depending on the levels and concentrations in this reaction you may get the sodium to disolve a little otherwise nothing will happen
What would happen to the rate of a reaction with rate law rate k NO2H2 if the concentration of NO were doubled?
The rate would be one-fourth. Correct on Apex.
What would happen to the area of a triangle when both dimensions are doubled?
If both dimensions are doubled then the area is quadrupled. This is true of any geometric shape.
What would happen to the rate of a reaction with rate law rate k NO2H2 if NO were halved?
In the rate law given as rate = k[NO2][H2], the concentration of NO does not appear, so the rate of the reaction is independent of its concentration. Therefore, if the concentration of NO were halved, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged as long as the concentrations of NO2 and H2 remain constant.
What would happen if the amount of molecules in earth's atmosphere doubled?
They would create a double greenhouse effect
What would happen to the rate of a reaction with rate law kNO2H2 if the concentration of NO were doubled?
The rate would be four times larger
When sodium hydroxide dissolve in water estimate the temperature change when the amount sodium hydroxide is doubled?
When sodium hydroxide dissolves in water, it undergoes an exothermic reaction, releasing heat. If the amount of sodium hydroxide is doubled, the same amount of heat will be released, resulting in an increase in temperature. The specific temperature change would depend on the initial amounts and concentrations of the sodium hydroxide and water.
Which equation would be used to calculate the rate constant from inital concentrations?
To calculate the rate constant (k) from initial concentrations, you would typically use the rate law equation for the reaction, which is expressed as ( \text{Rate} = k[A]^m[B]^n ), where ( [A] ) and ( [B] ) are the initial concentrations of the reactants, and ( m ) and ( n ) are their respective reaction orders. By measuring the initial rate of the reaction and substituting the initial concentrations into the rate law, you can rearrange the equation to solve for the rate constant ( k ).
What would happen to the volume of a rectangular prism if its side lengths were doubled?
The volume would increase by a factor of 23 = 8
What would happen to the rate of a reaction with rate law rate k NO2H2 if the concentration of H2 were halved?
If the concentration of H2 is halved, it would also halve the rate of the reaction, assuming H2 is a reactant in the rate law. This is because the rate law is directly proportional to the concentrations of reactants.
