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kolezko [41]
3 years ago
10

For the hypothetical compound AX2, which of the following statements is true?. a.) A is a nonmetal from Group 3A, and X is a met

al from Group 1A.. . b.) A is a metal from Group 1A, and X is a metal from Group 7A.. . c.) A is a metal from Group 3A, and X is a nonmetal from Group 5A.. . d.) A is a metal from Group 2A, and X is a nonmetal from Group 7A..
Chemistry
2 answers:
Mekhanik [1.2K]3 years ago
7 0

Answer:

answer is d

Explanation:

i did the test so i know

Ierofanga [76]3 years ago
4 0

The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X  belonging to Group 7A with -1 charge. Answer is D.
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ruslelena [56]

Answer:

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Explanation:

<em>The dissolution of HCl is  HCl → H⁺ + Cl⁻</em>

  • To solve part A) we need to calculate the concentration of H⁺, to do that we need the moles of H⁺ and the volume.

The problem gives us V=2.5 L, and the moles can be calculated using the molecular weight of HCl, 36.46 g/mol:

0.35g_{HCl}*\frac{1mol_{HCl}}{36.46g_{HCl}} *\frac{1molH^{+}}{1mol HCl} = 9.60*10⁻³ mol H⁺

So the concentration of H⁺ is

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pH = -log [H⁺] = -log (3.84 * 10⁻³) = 2.42

  • <em>The dissolution of NaOH is  NaOH → Na⁺ + OH⁻</em>
  • Now we calculate [OH⁻], we already know that V = 2.0 L, and a similar process is used to calculate the moles of OH⁻, keeping in mind the molecular weight of NaOH, 40 g/mol:

0.80g_{NaOH}*\frac{1mol_{NaOH}}{40g_{NaOH}} *\frac{1molOH^{-}}{1mol NaOH}= 0.02 mol OH⁻

[OH⁻] = 0.02 mol / 2.0 L = 0.01

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Answer:

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