Answer:
The formula of the original halide is SrCl₂.
Explanation:
- The balanced equation of this reaction is:
SrX₂ + H₂SO₄ → SrSO₄ + 2 HX, where X is the halide.
- From the equation stichiometry, 1.0 mole of strontium halide will result in 1.0 mole of SrSO₄.
- The number of moles of SrSO₄ <em>(n = mass/molar mass) </em>= (0.755 g) / (183.68 g/mole) = 4.11 x 10⁻³ mole.
- The number of moles of SrX are 4.11 x 10⁻³ moles from the stichiometry of the balanced equation.
- n = mass / molar mass, n = 4.11 x 10⁻³ moles and mass = 0.652 g.
- The molar mass of SrX₂ = mass / n = (0.652) / (4.11 x 10⁻³ moles) = 158.62 g/mole.
- The molar mass of SrX₂ (158.62 g/mole) = Atomic mass of Sr (87.62 g/mole) + (2 x Atomic mass of halide X).
- The atomic mass of halide X = (158.62 g/mole) - (87.62 g/mole) / 2 = 71 / 2 g/mole = 35.5 g/mole.
- This is the atomic mass of Cl.
- <em>So, the formula of the original halide is SrCl₂</em>.
Answer:
B
Explanation:
Having the same electronegativity means that the both elements will form a nonpolar bond. The shared electrons of the bind will be found equidistant from the nucleus of the two atoms because none has an unusual ability to attract the electrons of a bond towards itself.
Answer:
5 moles of NO₂ will remain after the reaction is complete
Explanation:
We state the reaction:
3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g)
3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:
If 3 moles of nitric oxide need 1 mol of water to react
Then, 26 moles of NO₂ may need (26 .1) / 3 = 8.67 moles of H₂O
We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:
1 mol of water needs 3 moles of oxide to react
Therefore, 7 moles of water will need (7 .3)/1 = 21 moles of oxide
We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains
Answer:
pH = 12.15
Explanation:
To determine the pH of the HCl and KOH mixture, we need to know that the reaction is a neutralization type.
HCl + KOH → H₂O + KCl
We need to determine the moles of each compound
M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl
M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH
The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl
HCl + KOH → H₂O + KCl
3 m 4 m -
1 m 3 m
As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.
1 mmol of KOH has 1 mmol of OH⁻, so the [OH⁻] will be 1 mmol / Tot volume
[OH⁻] 1 mmol / 70 mL = 0.014285 M
- log [OH⁻] = 1.85 → pH = 14 - pOH → 14 - 1.85 = 12.15
