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Novay_Z [31]
3 years ago
9

Write complete ionic equation to show the reaction of aqueous hg2(no3)2 with aqueous sodium chloride to form solid hg2cl2 and aq

ueous sodium nitrate. express your answer as a chemical equation. identify all of the phases in your answer.
Chemistry
2 answers:
Naddik [55]3 years ago
6 0

First we start with our molecular equation to ensure all atoms are balanced in order to maintain the law of conservation of mass.

Hg_2(NO_3)_2_(_a_q_) + NaCl_(_a_q_) \implies Hg_2Cl_2_(_s_) + NaNO_3_(_a_q_).

The equation is unbalanced since we have 1 Cl atom on left hand side as opposed to 2 on the right hand side and we have 2 NO_3 molecules on left hand side and 1 on the right hand side. So we add a coefficient of 2 on  both the NaCl and NaNO_3

Hg_2(NO_3)_2_(_a_q_) + 2NaCl_(_a_q_) \implies Hg_2Cl_2_(_s_) + 2NaNO_3_(_a_q_)

The reaction occurs between solutes that are soluble in water to give a solid as a precipitate. So the complete ionic equation will show the soluble compounds in their dissociated states as ions in water and the insoluble compound as a solid.

2Hg^+_(_a_q_) + 2NO_3^-_(_a_q_) + 2Na^+_(_a_q_) + 2Cl^- _(_a_q_) \implies Hg_2Cl_2_(_s_) + 2Na^+_(_a_q_) + 2NO_3^-_(_a_q_)

Andrews [41]3 years ago
5 0
<span>Answer: Hg2(NO3)2 + 2 NaCl --------> Hg2Cl2 + 2 NaNO3 Hg2+2 + 2NO3-1 + 2Na+ Cl- ----> Hg2Cl</span>
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3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
Citrus2011 [14]

<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

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3 years ago
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Mazyrski [523]

Answer:

[H₃O⁺] = 0.05 M & [OH⁻] = 2.0 x 10⁻¹³.

Explanation:

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<em>HNO₃ + H₂O → H₃O⁺ + NO₃⁻.</em>

  • The concentration of hydronium ion is equal to the concentration of HNO₃:

[H₃O⁺] = 0.05 M.

∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.

<em>∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] </em>= 10⁻¹⁴/0.05 = <em>2.0 x 10⁻¹³.</em>

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