Ammonium iodide, or NH₄I, is a salt. Since all salts are strong electrolytes, when ammonium iodide is dissolved in water, it would dissociate into ammonium ions and iodide ions. Hence, the major species present would be: <em>NH₄⁺ and I⁻ ions</em>.
Explanation:
P1V1 = P2V2
(100.7 kPa)(0.75 L) = (99.8 kPa)V2
V2 = (100.7 kPa)(0.75 L)/(99.8 kPa)
= 0.757 L
Answer:
Explanation:
6.90 mol x grams
2As + 6NaOH = 2Na3AsO3 + 3H2
6 mol 2 mol
192 g/mol
6.90 mol NaOH x 2 mol Na3AsO3/6 mol NaOH== 2.3 mol Na3AsO3
2.3 mol Na3AsO3 x 192 g/mol = 442 g Na3AsO3
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