<span>1. MgBr2
Soluble.
Rule: all the binary compounds of the group 17 (different to F) with metals are solubles, except those formed with Ag, Hg (I) and Pb.
2. PbI2
Insoluble.
Rule: it is one of the exceptions stated in the rule above.
3. (NH4)2CO3
Soluble.
Rule: salts containing NH4(+) are soluble.
4. ZnSO4
Soluble
Rule: </span><span>This salt is not an exception to the rule that most sulfate salts are soluble. Important exceptions to this rule include BaSO4,
PbSO4, Ag2SO4 and SrSO4
5. Sr(OH)2
Soluble (slightly soluble).
Rule: </span><span>Hydroxide salts of Group II elements (Ca,
Sr, and Ba) are slightly soluble</span>
Explanation:
To remove the salt from the oil, I will add water to dissolve the salt from it.
Oil is an organic molecule that is non-polar
Salt is polar ionic compound
Salt will not dissolve in the oil.
- Take the mixture.
- Add water to it.
- Water and oil are immiscible
- Shake the new heterogeneous mixture vigorously.
- leave to settle.
- Oil will come on top of the water.
- You can skim off the oil layer on top.
- Then heat the water and salt solution.
- This leaves the oil behind.
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Answer:
6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O
Explanation:
First, we need to write the half-reactions:
2Br⁻ → Br₂ + 2e⁻ Oxidation -Balanced yet-
XeO₃ → Xe Reduction
To balance the reduction in acidic aqueous solution we need to add waters in the other side of the reaction as oxygens are present:
XeO₃ → Xe + 3H₂O
And H⁺ as hydrogens from water we have:
XeO₃ + 6H⁺ → Xe + 3H₂O
To balance the charge:
<h3>XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O Reduction -Balanced-</h3><h3 />
To cancel out the electrons of both half-reaction we need to multiply oxidation 3 times:
6Br⁻ → 3Br₂ + 6e⁻
XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O
And the balanced reaction in acidic aqueous solution is the sum of both half-reactions:
<h3>6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O </h3>
Answer:
Wavelength is 10m
Explanation:
To solve this problem, we use the formula for wavelength
Wavelength = wave velocity / frequency
Wavelength = 50m/s / 5Hz
= 10m