Answer:
See explanation
Explanation:
Using the formula
°C = (F-32) × 5/9
Where;
°C = temperature in degrees centigrade
F= temperature in Fahrenheit
F= (9/5 ×°C) +32
F= (9/5 × 110) + 32
F= 230°F
To convert -78°C to Kelvin
-78°C + 273 = 195 K
Answer:
4.7 g. Option 5 is the right one.
Explanation:
4Al(s) + 3O₂ (g) ⇄ 2Al₂O₃ (s)
We convert the mass of reactants to moles, in order to determine the limiting.
2.5 g Al / 26.98 g/mol → 0.092 moles of Al
2.5 g O₂ / 32g/mol → 0.078 moles of O₂
Ratio is 4:3. 4 moles of Al react with 3 moles of O₂
Then, 0.092 moles of O₂ would react with (0.092 . 3)/ 4 = 0.069 moles O₂
We have 0.078 moles of O₂ and we need 0.069 moles, the oxygen is the limiting in excess. Therefore the Al is the limiting reactant.
Ratio is 4:2. 4 moles of Al, can produce 2 moles of Al₂O₃
Then, 0.092 moles of Al would produce (0.092 .2) / 4 = 0.046 moles
If we convert the moles to mass, we find the anwer:
0.046 mol . 101.96 g/mol = 4.69 g
Answer: the valence electron for phosphorus is 5. To achieve an octet electron arrangement, it needs to lose 5 electrons or gain 3 electrons. It is easier to gain 3 electrons than to lose 5 electrons. So phosphorus has to gain 3 electrons.
Explanation:
Hope it helps sorry if it doesn't
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
Answer:
Classification will be Potassium, Bromine, and Argon
Explanation:
- Potassium is more likely to lose electrons and form positive ion
- Bromine actually gain electrons and forms negative ion
- Argon does not lose or gain electrons
