This is false because it doesn't matter about the mass
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:
C. spontaneous at all temperatures
Explanation:
The spontaneity of reaction is determined by the sign of the gibbs free energy.
A negative sign denotes that the reaction is spontaneous, positive sign means the reaction is not spontaneous.
From the question;
ΔS° = +253 J/K
ΔH° = -125 kJ/mol
ΔG = ΔH° - TΔS°
From the data given, the condition in which we can obtain a negative value of G, is at any value of T.
For any value of T, G would always be a negative value.
This means the correct option is option C.
<span>For this reaction, oxidation number of Carbon in
CO would be +2 while oxidation number of carbon in CO2 would be +4 and so this
means that carbon has oxidized. Oxidation number of nitrogen in NO is +2. While
oxidation number of nitrogen in N2 is 0 so this means that nitrogen had reduced.
The reducing agent is the one which provides electrons by oxidizing itself so
in this case; CO is the reducing agent while the C in CO oxidized to produce
electrons. </span><span>I
am hoping that this answer has satisfied your query about and it will be able
to help you, and if you’d like, feel free to ask another question.</span>
