Answer:
The answer to your question is 0.5 liters
Explanation:
Data
[CaCl₂] = 4.0 M
number of moles = 2
volume = ?
Process
To solve this problem use the formula of Molarity and solve it for volume, substitute the values and simplify.
-Formula
Molarity = moles / volume
-Solve for volume
Volume = moles / molarity
-Substitution
Volume = 2/4
-Simplification
Volume = 0.5 liters.
The symbol %v/v means percent by volume. Assuming there is no volume effects when these substances are mixed, we calculate as follows:
% v/v = (25 mL ethanol / 25 mL + 150 mL ) x 100
%v/v = 14.29 mL ethanol / mL solution
Hope this answers the question.
1) Chemical equation
Na2 SiO3 (s) + 8 HF (aq) ---> H2 Si F6 (aq) + 2 Na F (aq) + 3H2O (l)
It is balanced
2) Molar ratios
1 mol Na2 SiO3 : 8 mol HF.
3) Proportion
0.340 mol Na2 SiO3 * 8 mol HF / 1mol Na2SiO3 = 2.72 mol HF.
Answer: 2.72 mol HF
Explanation:
Common ion effect is defined as the effect which occurs on equilibrium when a common ion (an ion which is already present in the solution) is added to a solution. This effect generally decreases the solubility of a solute.
Equilibrium reaction of strontium sulfate and sodium sulfate follows the equation:


According to Le-Chateliers principle: If there is any change in the variables of the reaction, the equilibrium will shift in the direction in order to minimize the effect.
In the equilibrium reactions, hypochlorite ion is getting increased on the product side, so the equilibrium will shift in the direction to minimize this effect, which is in the direction of hydrogen hypochlorite.
Thus, the addition hypochlorite ions will shift the equilibrium in the left direction.
The dissociation of hydrogen hypochlorite is suppressed due to the common ion effect.
Answer:
(a)

(b)

Explanation:
Hello,
(a) In this case, as the reaction is second-ordered, one uses the following kinetic equation to compute the concentration of NOBr after 22 seconds:
![\frac{1}{[NOBr]}=kt +\frac{1}{[NOBr]_0}\\\frac{1}{[NOBr]}=\frac{0.8}{M*s}*22s+\frac{1}{0.086M}=\frac{29.3}{M}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3Dkt%20%2B%5Cfrac%7B1%7D%7B%5BNOBr%5D_0%7D%5C%5C%5Cfrac%7B1%7D%7B%5BNOBr%5D%7D%3D%5Cfrac%7B0.8%7D%7BM%2As%7D%2A22s%2B%5Cfrac%7B1%7D%7B0.086M%7D%3D%5Cfrac%7B29.3%7D%7BM%7D%5C%5C)
![[NOBr]=\frac{1}{29.2/M}=0.0342M](https://tex.z-dn.net/?f=%5BNOBr%5D%3D%5Cfrac%7B1%7D%7B29.2%2FM%7D%3D0.0342M)
(b) Now, for a second-order reaction, the half-life is computed as shown below:
![t_{1/2}=\frac{1}{k[NOBr]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BNOBr%5D_0%7D)
Therefore, for the given initial concentrations one obtains:

Best regards.