5.27x10(24) superscript atoms CHC13
Answer:
Pb⁺²(aq) + CO₃⁻²(aq) → PbCO₃ (s)
In net ionic equation we cancel the ions that have equal moles on both sides so Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.
Explanation:
Net ionic equation:
In net ionic equation we only write the ions that are involved in reaction. If the system have same moles of ions in initial and final stages we cancel them as they have the same amount and are present in ionic form in the reaction medium. To formulate an ionic equation we just cancel the ions which have the same moles in initial and final stages.
Chemical equation:
Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + NaNO₃ (aq)
Balanced chemical equation:
In a balanced chemical equation we write the reactants and products in molecular form with number of moles.
Pb(NO₃)₂ (aq) + Na₂CO₃(aq) → PbCO₃ (s) + 2NaNO₃ (aq)
Ionic equation:
In ionic equation we write the equation in ionic form. It involves all the ions which will produce when we add any ionic compound in reaction medium.
Pb⁺² +2NO₃⁻¹ + CO₃⁻² + 2Na⁺¹ → PbCO3 (s) + 2NO₃⁻¹ (aq) + 2Na⁺¹ (aq)
Net ionic equation
In net ionic equation we cancel the ions that have equal moles on both sides. As we can see in the above ionic equation that Na⁺¹ and NO₃⁻¹ have equal moles on both sides so we canceled them.
Pb⁺²(aq) + CO₃⁻²(aq) → PbCO₃ (s)
Answer: any answer choices
Explanation:
The required volume of water to make the dilute solution of 0.5 M is 188 mL.
<h3>How do we calculate the required volume?</h3>
Required volume of water to dilute the stock solution will be calculated by using the below equation as:
M₁V₁ = M₂V₂, where
- M₁ & V₁ are the molarity and volume of stock solution.
- M₂ & V₂ are the molarity and volume of dilute solution.
On putting values from the question to the above equation, we get
V₂ = (2)(47) / (0.5) = 188mL
Hence required volume of water is 188 mL.
To know more about volume & concentration, visit the below link:
brainly.com/question/7208546
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Answer:
d
Explanation:
its a proposed explanation based off limited evidence.
