Answer:
- <em>2NaCl → 2Na + Cl₂, ΔH = 822 kJ </em>
Explanation:
The chemical <em>equation</em> for the <em>formation of NaCl</em> is:
- Na + (1/2) Cl₂ → NaCl , ΔH = - 411 kJ
That equation means that 1 mole of NaCl is formed by the reaction of 1 mole of Na and 1/2 mole of Cl₂, with a release of energy of 411 kJ.
The <em>decomposition</em> of <em>NaCl</em> is the inverse of the <em>formation</em> reaction; thus, you swift products and reactants and inverse the sign of the <em>change in enthalpy:</em>
- NaCl → Na + 1/2 Cl₂, ΔH = 411 kJ
Since you want the decomposition of 2 moles you multiply the equation and the ΔH by 2:
- 2NaCl → 2Na + Cl₂, ΔH = 822 kJ ← answer
Answer:
lake water, swamp water, like, any dirty and unpurified forms of water is hard water
Explanation:
Answer:
1) high pH is required
2) other ions are precipitated along with the strontium ions
Explanation:
According to the solubility rules all phosphates are insoluble except those of sodium, potassium, and ammonium. This implies that strontium phosphate is insoluble in water. This explains why strontium ions can be precipitated from drinking water supply using phosphate. The main problem with the precipitation of strontium using phosphate is that it usually requires a high pH as the precipitation occurs under very alkaline conditions.
The main reason why the results may not be accurate is that other ions are precipitated along with the strontium such as calcium ions and magnesium ions. This may lead to inaccurate determination of the amount of strontium ions present.