One mole of Na2SO4 is 6.022 * 10^(23) molecules. We can divide this into the quantity in the question to find a value of 1.5/6.022 = 0.2491 moles. Rounded to two significant figures and put in scientific notation, we can rewrite this quantity as 2.5 * 10^(-1) moles
The answer is 1.164451827 moles
Answer:
The Equilibrium constant K is far greater than 1; K>>1
Explanation:
The equilibrium constant, K, for any given reaction at equilibrium, is defined as the ratio of the concentration of the products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.
It tells us more about how how bigger or smaller the concentration of products is to that of the reactants when a reaction attains equilibrium. From the given data, as the color of the reactant mixture (Br2 is reddish-brown, and H2 is colourless) fades, more of the colorless product (HBr is colorless) is being formed as the reaction approaches equilibrium. This indicates yhat the concentration of products becomes relatively higher than that of the reactants as the reaction progresses towards equilibrium, the equilibrium constant K, must be greater than 1 therefore.
Answer:
0.758 V.
Explanation:
Hello!
In this case, case when we include the effect of concentration on an electrochemical cell, we need to consider the Nerst equation at 25 °C:

Whereas n stands for the number of moles of transferred electrons and Q the reaction quotient relating the concentration of the oxidized species over the concentration of the reduced species. In such a way, we can write the undergoing half-reactions in the cell, considering the iron's one is reversed because it has the most positive standard potential so it tends to reduction:

It means that the concentration of the oxidized species is 0.002 M (that of nickel), that of the reduced species is 0.40 M and there are two moles of transferred electrons; therefore, the generated potential turns out:

Beat regards!
Answer : The balanced equations will be:

Explanation :
The general rate of reaction is,

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
![\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20A%7D%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20B%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20C%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20D%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
![Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
Now we have to determine the balanced equations corresponding to the following rate expressions.
![Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BCH_4%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BCO_2%5D%7D%7Bdt%7D)
The balanced equations will be:
