Answer:
0.25
Explanation:
volume= number of moles over concentration
hence v=1.5/6
Answer:
1.7 mL
Explanation:
<em>A chemist must prepare 550.0 mL of hydrochloric acid solution with a pH of 1.60 at 25 °C. He will do this in three steps: Fill a 550.0 mL volumetric flask about halfway with distilled water. Measure out a small volume of concentrated (8.0 M) stock hydrochloric acid solution and add it to the flask. Fill the flask to the mark with distilled water. Calculate the volume of concentrated hydrochloric acid that the chemist must measure out in the second step. Round your answer to 2 significant digits.</em>
Step 1: Calculate [H⁺] in the dilute solution
We will use the following expresion.
pH = -log [H⁺]
[H⁺] = antilog - pH = antilog -1.60 = 0.0251 M
Since HCl is a strong monoprotic acid, the concentration of HCl in the dilute solution is 0.0251 M.
Step 2: Calculate the volume of the concentrated HCl solution
We want to prepare 550.0 mL of a 0.0251 M HCl solution. We can calculate the volume of the 8.0 M solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂/C₁
V₁ = 0.0251 M × 550.0 mL/8.0 M = 1.7 mL
The volume that will be occupied at 735 torr and 57 c is 23.12 L
<u><em>calculation</em></u>
- <u><em> </em></u> At STP temperature=273 k and pressure=760 torr
- <u><em> </em></u>by use of combined gas formula
that is P1V1/T1= P2V2/T2
where; P1 =760 torr
T1= 273 K
V1= 18.5 L
P2= 735 torr
T2= 57+273= 330 K
V2=?
- by making V2 the formula of subject
V2= T2P1V1/P2T1
V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L
Explanation:
Equilibrium constant of reaction = 
Concentration of NO = ![[NO]=\frac{2.69\times 10^{-2} mol}{1 L}=2.69\times 10^{-2} M](https://tex.z-dn.net/?f=%5BNO%5D%3D%5Cfrac%7B2.69%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D2.69%5Ctimes%2010%5E%7B-2%7D%20M)
Concentration of bromine gas = ![[Br_2]=\frac{3.85\times 10^{-2} mol}{1 L}=3.85\times 10^{-2} M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7B3.85%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D3.85%5Ctimes%2010%5E%7B-2%7D%20M)
Concentration of NOBr gas = ![[Br_2]=\frac{9.56\times 10^{-2} mol}{1 L}=9.56\times 10^{-2} M](https://tex.z-dn.net/?f=%5BBr_2%5D%3D%5Cfrac%7B9.56%5Ctimes%2010%5E%7B-2%7D%20mol%7D%7B1%20L%7D%3D9.56%5Ctimes%2010%5E%7B-2%7D%20M)
The reaction quotient is given as:
![Q=\frac{[NOBr]^2}{[NO]^2[Br_2]}=\frac{(9.56\times 10^{-2} M)^2}{(2.69\times 10^{-2} M)^2\times 3.85\times 10^{-2} M}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BNOBr%5D%5E2%7D%7B%5BNO%5D%5E2%5BBr_2%5D%7D%3D%5Cfrac%7B%289.56%5Ctimes%2010%5E%7B-2%7D%20M%29%5E2%7D%7B%282.69%5Ctimes%2010%5E%7B-2%7D%20M%29%5E2%5Ctimes%203.85%5Ctimes%2010%5E%7B-2%7D%20M%7D)
The reaction will go in backward direction in order to achieve an equilibrium state.
1. In order to reach equilibrium NOBr (g) must be produced. False
2. In order to reach equilibrium K must decrease. False
3. In order to reach equilibrium NO must be produced. True
4. Q. is less than K . False
5. The reaction is at equilibrium. No further reaction will occur. False
To calculate the pH of a solution that has a [H3O+] of 7.22x10^-7. You would do the following
pH=-log[H3O+]
pH=-log[7.22x10^-7]
pH=?
