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2 years ago

List the spectator ions in the following reaction.

Chemistry

1 answer:

Rudik [331]2 years ago

4 0

Answer:

2Na⁺ (aq) and 2OH⁻(aq)

Explanation:

Spectator ions:

Spectator ions are those ions which are same on both side of chemical reaction. These ions are same in the reactant side and product side. Their presence can not effect the chemical equilibrium that's why when we write the net ionic equation these ions are neglect or omitted.

Given ionic equation:

Ba⁺²(aq) + 2OH⁻(aq) + 2Na⁺ (aq) + CO²⁻₃(aq) → BaCO₃(s) + + 2Na⁺ (aq) + 2OH⁻(aq)

In given ionic equation by omitting the spectator ions i.e, 2Na⁺ (aq) and 2OH⁻(aq) net ionic equation can be written as,

Net ionic equation:

Ba⁺²(aq) + CO²⁻₃(aq) → BaCO₃(s)

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Can someone help me?

kupik [55]

Answer:

#1)2.23hrs

Explanation:

= 134/60

=2.23

=2.23hrs

4 0

3 years ago

A small, hard-shelled fossil that resembles a modern-day ocean organism was found

Paul [167]

Answer:

The desert was one covered by ocean water.

Explanation:

5 0

2 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

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

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

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

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

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

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

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago

The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp

EleoNora [17]

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

6 0

3 years ago

Consider the following reaction: C6H6 + O2 \longrightarrow ⟶ CO2 + H2O 39.7 grams of C6H6 are allowed to react with 105.7 g of O

Ivenika [448]

Answer:

116.3 grCO2

Explanation:

1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side

C6H6 +15/2 O2⟶ 6CO2 +3 H2O

2nd - we calculate the limiting reagent

39.2gr C6H6*(240grO2/78grC6H6)=120 grO2

we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent

3rd - we use the limiting reagent to calculate the amount of CO2 in grams

105.7grO2*(264grCO2/240grO2)=116.3 grCO2

7 0

2 years ago