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3 years ago

Consider the total ionic equation below.

Chemistry

2 answers:

Tpy6a [65]3 years ago

4 0

Answer:

The net ionic equation is

Ba²⁺ + CO₃²⁻ → BaCO₃(s)

Explanation:

The chemical reaction described is given as

Ba²⁺ + 2NO₃⁻ + 2Na⁺ + CO₃²⁻ → BaCO₃(s) + 2Na⁺ + 2NO₃⁻

The net ionic equation is obtained by eliminating the ions that appear on both sides of the equation.

And those ions are 2Na⁺ and 2NO₃⁻.

This is a double displacement reaction in which BaNO₃ (aq) reacts with Na₂CO₃ (aq) to form BaCO₃ (s) and NaNO₃ (aq).

The aqueous solutions exist in ionic forms in the reaction and in the net ionic equation, ioms that appear on both sides of the overall equation obviously cancel out.

Hope this Helps!!!

Alexeev081 [22]3 years ago

4 0

Answer:

The correct answer is Ba2+ +2NO-3 =Ba(NO3)2

Explanation:

did it on edu

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Acetaldehyde decomposes at 750 K: CH3CHO → CO + CH4. The reaction is first order in acetaldehyde and the half-life of the reacti

Degger [83]

Answer:

k = 1.3 x 10⁻³ s⁻¹

Explanation:

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4 0

3 years ago

Salt in crude oil must be removed before the oil undergoes processing in a refinery. The

irina1246 [14]

Answer:

$\large \boxed{0.64 \, \%}$

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

$\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}$

2. Calculate the mass of salt in the salty oil

$\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}$

3. Calculate the mass of salt in the spent water

$\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}$

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g

5. Concentration of salt in washed oil

$\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}$