Answer:
0.15 moles of sodium hydroxide are in the solution.
Explanation:
Molarity is a unit for expressing concentration of solutions. Molarity is defined as the number of moles of solute per liter of solution.
The molarity of a solution is calculated by dividing the moles of the solute by the liters of the solution:
Molarity is expressed in units (
).
So, a molarity of 0.750 M indicates that 0.750 moles are present in 1 L of solution. Then the following rule of three can be applied: if in 1000 mL (being 1 L = 1000 mL) there are 0.750 moles, in 200 mL how many moles are there?
amount of moles=0.15
<u><em>
0.15 moles of sodium hydroxide are in the solution.</em></u>
<u>Answer:</u>
<u>For a:</u> The equilibrium mixture contains primarily reactants.
<u>For b:</u> The equilibrium mixture contains primarily products.
<u>Explanation:</u>
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium.
For the given chemical reactions:
The chemical equation follows:
The expression of 
for above reaction follows:
![K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCN%5E-%5D%5BH_3O%5E%2B%5D%7D%7B%5BHCN%5D%5BH_2O%5D%7D%3D6.2%5Ctimes%2010%5E%7B-10%7D)
As, , the reaction will be favored on the reactant side.
Hence, the equilibrium mixture contains primarily reactants.
The chemical equation follows:
The expression of for above reaction follows:
![K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHCl%5D%5E2%7D%7B%5BH_2%5D%5BCl_2%5D%7D%3D2.51%5Ctimes%2010%5E%7B4%7D)
As, , the reaction will be favored on the product side.
Hence, the equilibrium mixture contains primarily products.
Answer: 3.57M
Explanation:Please see attachment for explanation
Answer:
The partial pressure of argon in the jar is 0.944 kilopascal.
Explanation:
Step 1: Data given
Volume of the jar of air = 25.0 L
Number of moles argon = 0.0104 moles
Temperature = 273 K
Step 2: Calculate the pressure of argon with the ideal gas law
p*V = nRT
p = (nRT)/V
⇒ with n = the number of moles of argon = 0.0104 moles
⇒ with R = the gas constant = 0.0821 L*atm/mol*K
⇒ with T = the temperature = 273 K
⇒ with V = the volume of the jar = 25.0 L
p = (0.0104 * 0.0821 * 273)/25.0
p = 0.00932 atm
1 atm =101.3 kPa
0.00932 atm = 101.3 * 0.00932 = 0.944 kPa
The partial pressure of argon in the jar is 0.944 kilopascal.
It would be d bc when it’s at its lowest point it means not that much energy and that would be its lowest point which is D
