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EleoNora [17]
3 years ago
11

Consider the following ionic compounds: CdCO3, Na2S, PbSO4, (NH4)3PO4, and Hg2Cl2. Which compounds will be soluble when added to

water
Chemistry
1 answer:
k0ka [10]3 years ago
7 0

Answer:

Na2S, (NH4)3PO4

Explanation:

We can decide what compounds are soluble by considering the solubility rules that apply.

CdCO3 is not soluble in water because all carbonates are insoluble except those of ammonium, sodium and potassium.

Na2S is soluble in water because all sulphides are insoluble except those of sodium, potassium and ammonium.

PbSO4 is insoluble in water because all sulphates are soluble except those of lead and barium. The sulphate of calcium is only slightly soluble in water.

(NH4)3PO4 is soluble in water because all phosphates are insoluble except those of sodium, potassium and ammonium.

Hg2Cl2 is insoluble in water because all chlorides are soluble except those of lead, mercury II and silver.

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Through electrolysis, a student collects 6.6 g of hydrogen gas and 52.4 g of oxygen gas. The reaction was allowed to proceed to
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Answer: 59 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2H_2O\rightarrow 2H_2+O_2

Given: mass of hydrogen = 6.6 g

mass of oxygen = 52.4 g

Mass of products = Mass of hydrogen + mass of oxygen = 6.6 +52.4 = 59 g grams

Thus mass or reactant = mass of water

Mass of reactants = mass of products = 59 g

Thus the mass of water initially present was 59 g.

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If C and Cl combined to make a compound, would it be covalent or ionic?
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Answer:

it would be a covalent bond

Explanation:

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6 0
3 years ago
write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi
diamong [38]

Answer:

First ionization of lithium:

\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}.

Second ionization of lithium:

\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}.

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

  • Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol (g) next to any atom or ion in the equation.
  • The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
  • Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

  • The products shall contain a gaseous ion with charge +1 \text{Li}^{+}\;(g) as well as an electron \text{e}^{-}.
  • Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: \text{Li}\;(g).
  • Hence the equation: \text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}.

Second Ionization Energy of Li:

  • The product shall contain a gaseous ion with charge +2: \text{Li}^{2+}\;(g) as well as an electron \text{e}^{-}.
  • Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?
  • The equation for this process is \text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}.
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How many electrons elements are present in period 2 ?​
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