Antimony has two naturally occurring isotopes. Their abundance is given in the pic attached below
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B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.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.
Th actual yield of the reaction is 24.86 g
We'll begin by calculating the theoretical yield of the reaction.
2Na + Cl₂ → 2NaCl
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl from the balanced = 2 × 58.5 = 117 g
From the balanced equation above,
46 g of Na reacted to produce 117 g of NaCl.
Therefore,
11.5 g of Na will react to produce = (11.5 × 117) / 46 = 29.25 g of NaCl.
Thus, the theoretical yield of NaCl is 29.25 g.
Finally, we shall determine the actual yield of NaCl.
- Theoretical yield = 29.25 g
Actual yield = Percent yield × Theoretical yield
Actual yield = 85% × 29.25
Actual yield = 0.85 × 29.25 g
Actual yield = 24.86 g
Learn more about stoichiometry: brainly.com/question/25899385
Answer:
5.79 × 10^23 Oxygen atoms
Explanation:
Number of Oxygen atom in the compound = 4×3 = 12
Molar mass of Al2(SO4)3 = 342 g/mol.
No of mole = mass/molar mass = 2.74/342 = 8.01×10^-03 mole
2.74g of Al2(SO4)3 × 1 mole of Al2 (SO4)3 / 342g of Al2 (SO4)3 * 12 mole of Oxygen/ 1mole of Al(SO4)3 * 6.02×10^23 Oxygen atom/ 1 mole of Oxygen
= 5.79×10^23 Oxygen atoms