What is the formula for the compound formed by lead(lV) ions and chromate ions

(c) ammonium carbonate + lead nitrate ? Step-by-step solution Step 1 of 3 v Step 2 of 3 v Step 3 of 3 ^ c) The ions formed in so

kenny6666 [7]

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium carbonate and lead nitrate is given as:

$(NH_4)_2CO_3(aq.)+Pb(NO_3)_2(aq.)\rightarrow PbCO_3(s)+2NH_4NO_3(aq.)$

Ionic form of the above equation follows:

$2NH_4^+(aq.)+CO_3^{2-}(aq.)+Pb^{2+}(aq.)+2NO_3^-(aq.)\rightarrow PbCO_3(s)+2NH_4^+(aq.)+2NO_3^-(aq.)$

As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Pb^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow PbCO_3(s)$

Hence, the net ionic equation is written above.

3 0

2 years ago

What is a successful outcome for using a scientific method?

andrezito [222]

Developing a theory would be a successful outcome

3 0

2 years ago

The average dosage of oxcarbazepine for an epileptic child between the ages of 4 and 16 is 9.00 mg per 1 kg of body weight (9.00

melamori03 [73]

Convert the child weight (37.3 pounds) to kilograms

37.3 lb x 0.453 kg /1lb = "A kg"

multiply the dose (9.00mg/kg) by the weight of the child to find how much you need to give him

A kg * 9.00 mg/1kg = "B mg"

calculate the mL of suspension dividing the "B mg" by the concentration of the suspension 60.0 mg/mL

B mg * 1mL/ 60.0 mg = C mL oxcarbazepine

8 0

3 years ago

Read 2 more answers

Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of

Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

$\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles$