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GrogVix [38]
3 years ago
8

When aqueous solutions of sodium carbonate and zinc iodide are combined, solid zinc carbonate and a solution of sodium iodide ar

e formed. The net ionic equation for this reaction is:______
Chemistry
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:

CO₃²⁻ + Zn²⁺ → ZnCO₃(s)

Explanation:

The net ionic equation is defined as a chemical equation for a reaction that lists only those species participating in the reaction.

In the reaction of sodium carbonate (Na₂CO₃) with zinc iodide (ZnI₂) that produce solid zinc carbonate (ZnCO₃) and a solution of sodium iodide (NaI) the formula is:

2Na⁺ + CO₃²⁻ + Zn²⁺ + 2I⁻ → ZnCO₃(s) + 2Na⁺ + 2I⁻

The net ionic equation is:

<em> CO₃²⁻ + Zn²⁺ → ZnCO₃(s)</em>

<em></em>

I hope it helps!

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Why are pennies good for simulating radioactive atoms? Check all that apply. Pennies have two sides—one to represent radioactive
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Answer: all of those are the right choices. :) promise

Explanation:

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4 years ago
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a compound has a percent compostion of carbon equal to 48.8383%, hydrogen equal to 8.1636%, and oxygen equal to 43.1981%. what i
Margarita [4]

Answer:

C₂H₃O

Explanation:

From the question given above, the following data were obtained:

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

8 0
3 years ago
What are some HYDROPHILIC materials?
alisha [4.7K]
Including the alkanes, oils , fats and greasy substances in general
8 0
3 years ago
A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
Scilla [17]

Answer:

The essence including its given problem is outlined in the following segment on the context..

Explanation:

The given values are:

Moles of CO₂,

x = 0.01962

Moles of water,

\frac{y}{2} =0.01961

y=2\times 0.01961

  =0.03922

Compound's mass,

= 0.4647 g

Let the compound's formula will be:

C_{x}H_{y}O_{z}

Combustion's general equation will be:

⇒  C_{x}H_{y}O_{z}+x+(\frac{y}{4}-\frac{z}{2}) O_{2}=xCO_{2}+\frac{y}{2H_{2}O}

On putting the estimated values, we get

⇒  12\times x=1\times y+16\times z=0.4647

⇒  12\times 0.01962+1\times 0.03922+16\times z=0.4647

⇒  0.27466+16z=0.4647

⇒                     z=0.01187

Now,

x : y : z = 0.01962:0.03922:0.01187

           = \frac{0.01962}{0.0118}:\frac{0.03922}{0.0188}:\frac{0.0188}{0.0188}

           = 1.6:3.3:1.0

           = 3:6:2

So that the empirical formula seems to be "C₃H₆O₂".

8 0
3 years ago
An unknown compound contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 2.77 g sample of the compound
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The molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

<h3>Molecular formula of the compound</h3>

The molecular formula is calculated as follows;

CHNO  +  O₂ ------------> CO₂  + H₂O

Mass of carbon, C:  = (6.64 x 12)/44 = 1.81 g in 2.77 g sample

Mass of hydrogen, H: = (1.67 x 2)/18 = 0.186 g in 2.77 g sample

Mass of Nitrogen, N: = (2.77 x 0.143)/1.69 = 0.234 g

Mass of oxygen, O:  = 2.77 g - 1.81 g - 0.186 g - 0.234 g = 0.54 g

<h3>molar ratio of the elements: </h3>

C = 1.81 g = 0.15 mol

H = 0.186 g = 0.186 mol

N = 0.234 g = 0.017 mol

O = 0.54 g = 0.0337 mol

divide through with the smallest number of moles (0.017 mol);

C = 9

H = 11

N = 1

O = 2

Molecular formula = C₉H₁₁NO₂

Check the molar mass of the compound = (9 x 12) + (11 x 1) + (14) + (2 x 16) = 165 g/mol

Thus, the molecular formula of the unknown compound is determined as  C₉H₁₁NO₂.

Learn more about molecular mass here: brainly.com/question/21334167

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