Answer:
- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).
Explanation:
Hello,
In this case, since the van't Hoff factor is related with the species that result from the ionization of a chemical compound, we can see that that
- Aluminium sulfate Al2(SO4)3 dissociates in two aluminium ions and three sulfate ions, therefore, van't Hoff factor is 5 (incorrect).
- Ammonium nitrate NH4NO3 dissociates in one ammonium ions and one nitrate ion, therefore, van't Hoff factor is 2 (correct).
- Sodium sulfate Na2SO4 dissociates in two sodium ions and one sulfate, therefore, van't Hoff factor is 3 (correct).
- Sucrose is not ionized, therefore, van't Hoff factor is 1 (correct).
Best regards.
The percent yield of the reaction : 89.14%
Reaction of Ammonia and Oxygen in a lab :
4 NH₃ (g) + 5 O₂ (g) ⇒ 4 NO(g)+ 6 H₂O(g)
mass NH₃ = 80 g
mol NH₃ (MW=17 g/mol):
mass O₂ = 120 g
mol O₂(MW=32 g/mol) :
Mol ratio of reactants(to find limiting reatants) :
mol of H₂O based on O₂ as limiting reactants :
mol H₂O :
mass H₂O :
4.5 x 18 g/mol = 81 g
The percent yield :
Answer:
Heterogeneous Mixture. Have a good day! =)
Explanation:
B. ethyl alcohol
C.sulfuric acid
D.All chemicals are potentially dangerous
Answer:
D.All chemicals are potentially dangerous
Explanation:
No chemical is toxicologically neutral
Answer:
Explanation:
Hello,
In this case, since the result of the operation between two magnitudes is shown with the same significant figures of the shortest number, we obtain:
Next, we proceed as follows:
Nevertheless, since 1.012 is the shortest number and has four significant figures, the result is rounded to four significant figures, that is until the three but it rounded due to the fact that the next digit is five:
Regards.
Answer : The concentration of ion, pH and pOH of solution is, , 4.98 and 9.02 respectively.
Explanation : Given,
Concentration of ion =
pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.
The expression used for pH is:
First we have to calculate the pH.
The pOH of the solution is, 9.02
Now we have to calculate the pH.
The pH of the solution is, 4.98
Now we have to calculate the concentration.
The concentration is,
Answer:
pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978
Explanation:
Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C
As, pOH = - log [OH⁻]
⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022
The self-ionisation constant of water is given by
Kw = [H⁺] [OH⁻] and pKw = pH + pOH
Since, at room temperature (25°C): Kw = 1.0 × 10⁻¹⁴ and pKw = 14.
Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴
⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M
also,
pH + pOH = pKw = 14
⇒ pH = 14 - pOH = 14 - 9.022 = 4.978