Answer:
2 L
Explanation:
From the question given above, the following data were obtained:
Molarity of LiF = 2 M
Mole of LiF = 4 moles
Volume =?
Molarity of a solution is simply defined as the mole per unit litre of the solution. Mathematically, it is expressed as:
Molarity = mole / Volume
With the above formula, we can obtain the volume of the solution as shown below:
Molarity of LiF = 2 M
Mole of LiF = 4 moles
Volume =?
Molarity = mole / Volume
2 = 4 / volume
Cross multiply
2 × volume = 4
Divide both side by 2
Volume = 4/2
Volume = 2 L
Therefore, the volume of the solution is 2 L.
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:

2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Answer:
the first one to the third box
the second one to the fourth box
the third one to second box
and the fourth one to the first box
Explanation:
Percentage recovery gives us an idea of the amount of pure substance recovered after the chemical reaction. Percentage recovery can be more than 100 % or less than 100 %. Usually, in any experiment performed the weight percentage recovery will be less than 100. Percent recovery values greater than 100 show that the recovered compound is contaminated.
Amount of acetaminophen initially taken = 350 mg
Amount of acetaminophen obtained after recovery =185 mg

= 
= 52.9%