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Greeley [361]
2 years ago
14

Determine net ionic equations, if any, occuring when aqueous solutions of the following reactants are mixed. Select "True" or "F

alse" to indicate whether or not the stated reaction (or "no reaction") correctly corresponds to the expected observation in each case. Lead(II) nitrate and sodium chloride; No reaction occurs. Sodium bromide and hydrochloric acid; No reaction occurs. Nickel(II) chloride and lead(II) nitrate; Pb2+(aq) + 2Cl-(aq) --> PbCl2(s) Magnesium chloride and sodium hydroxide; No reaction occurs. Ammonium sulfate and barium nitrate; Ba2+(aq) + SO42-(aq) --> BaSO4(s)
Chemistry
1 answer:
nirvana33 [79]2 years ago
3 0

<u>Answer:</u>

<u>For 1:</u> The correct answer is False.

<u>For 2:</u> The correct answer is True.

<u>For 3:</u> The correct answer is True.

<u>For 4:</u> The correct answer is False.

<u>For 5:</u> The correct answer is True.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.  If no net ionic equation is formed, it is said that no reaction has occurred.

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. Solids, liquids and gases do not exist as ions.

  • <u>For 1:</u> Lead(II) nitrate and sodium chloride

The chemical equation for the reaction of lead (II) nitrate and sodium chloride is given as:

Pb(NO_3)_2(aq.)+2NaCl(aq.)\rightarrow PbCl_2(s)+2NaNO_3(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+2Na^+(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+2Na^+(aq.)+2NO_3^-(aq.)

As, sodium 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.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is False.

  • <u>For 2:</u> Sodium bromide and hydrochloric acid

The chemical equation for the reaction of sodium bromide and hydrochloric acid is given as:

NaBr(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+HBr(aq.)

Ionic form of the above equation follows:

Na^{+}(aq.)+Br^-(aq.)+H^+(aq.)+Cl^-(aq.)\rightarrow Na^+(aq.)+Cl^-(aq.)+H^+(aq.)+Br^-(aq.)

There are no spectator ions in the equation. So, the above reaction is the net ionic equation.

Hence, the correct answer is True.

  • <u>For 3:</u> Nickel (II) chloride and lead(II) nitrate

The chemical equation for the reaction of lead (II) nitrate and nickel (II) chloride is given as:

Pb(NO_3)_2(aq.)+NiCl_2(aq.)\rightarrow PbCl_2(s)+Ni(NO_3)_2(aq.)

Ionic form of the above equation follows:

Pb^{2+}(aq.)+2NO_3^-(aq.)+Ni^{2+}(aq.)+2Cl^-(aq.)\rightarrow PbCl_2(s)+Ni^{2+}(aq.)+2NO_3^-(aq.)

As, nickel 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.)+Cl^-(aq.)\rightarrow PbCl_2(s)

Hence, the correct answer is True.

  • <u>For 4:</u> Magnesium chloride and sodium hydroxide

The chemical equation for the reaction of magnesium chloride and sodium hydroxide is given as:

MgCl_2(aq.)+2NaOH(aq.)\rightarrow Mg(OH)_2(s)+2NaCl(aq.)

Ionic form of the above equation follows:

Mg^{2+}(aq.)+2Cl^-(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Mg(OH)_2(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride 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:

Mg^{2+}(aq.)+OH^-(aq.)\rightarrow Mg(OH)_2(s)

Hence, the correct answer is False.

  • <u>For 5:</u> Ammonium sulfate and barium nitrate

The chemical equation for the reaction of ammonium sulfate and barium nitrate is given as:

(NH_4)_2SO_4(aq.)+Ba(NO_3)_2(aq.)\rightarrow BaSO_4(s)+2NH_4NO_3(aq.)

Ionic form of the above equation follows:

2NH_4^{+}(aq.)+SO_4^{2-}(aq.)+Ba^{2+}(aq.)+2NO_3^-(aq.)\rightarrow BaSO_4(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:

Ba^{2+}(aq.)+SO_4^{2-}(aq.)\rightarrow BaSO_4(s)

Hence, the correct answer is True.

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How many grams of antifreeze C2H4(OH)2 would be required per 500 g of water to prevent the water from freezing at a temperature
Wewaii [24]

Answer:

333.7 g.

Explanation:

  • The depression in freezing point of water (ΔTf) due to adding a solute to it is given by: <em>ΔTf = Kf.m.</em>

Where, ΔTf is the depression in water freezing point (ΔTf = 20.0°C).

Kf is the molal freezing point depression constant of the solvent (Kf = 1.86 °C/m).

m is the molality of the solution.

<em>∴ m = ΔTf/Kf</em> = (20.0°C)/(1.86 °C/m) = <em>10.75 m.</em>

molaity (m) is the no. of moles of solute per kg of the solvent.

∵ m = (no. of moles of antifreeze C₂H₄(OH)₂)/(mass of water (kg))

∴ no. of moles of antifreeze C₂H₄(OH)₂ = (m)(mass of water (kg)) = (10.75 m)(0.5 kg) = 5.376 mol.

∵ no. of moles = mass/molar mass.

<em>∴ mass of antifreeze C₂H₄(OH)₂ = no. of moles x molar mass </em>= (5.376 mol)(62.07 g/mol) =<em> 333.7 g.</em>

5 0
2 years ago
When heating a flammable or volatile solvent for a recrystallization, which of these statements are correct? More than one answe
otez555 [7]

Explanation:

A volatile substance is defined as the substance which can easily evaporate into the atmosphere due to weak intermolecular forces present within its molecules.

Whereas a flammable substance is defined as a substance which is able to catch fire easily when it comes in contact with flame.

Hence, when we heat a flammable or volatile solvent for a recrystallization then it should be kept in mind that should heat the solvent in a stoppered flask to keep vapor away from any open flames so that it won't catch fire.

And, you should ensure that no one else is using an open flame near your experiment.

Thus, we can conclude that following statements are correct:

  • You should heat the solvent in a stoppered flask to keep vapor away from any open flames.
  • You should ensure that no one else is using an open flame near your experiment.
3 0
2 years ago
What is the molarity of a naoh solution if 11.9 ml of a 0.220 m h2so4 solution is required to neutralize a 25.0-ml sample of the
bulgar [2K]
0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule. 
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>

4 0
2 years ago
Read 2 more answers
Diborane, B2H6 a possible rocket propellant, can be made by using lithium hydride (LiH): 6 LiH+ 2 BCl2àB2H6+ 6 LiCl . If you mix
Genrish500 [490]

Answer :

(a) Limiting reactant = LiH

(b) The excess reactant = BCl_3

(c) The percent of excess reactant is, 50.87 %

(d) The percent yield of B_2H_6 or percent conversion of LiH to B_2H_6 is, 38.80 %

(e) The mass of LiCl produced is, 1066.42 lb

Explanation : Given,

Mass of LiH = 200 lb = 90718.5 g

conversion used : (1 lb = 453.592 g)

Mass of BCl_3 = 1000 lb = 453592 g

Molar mass of LiH = 7.95 g/mole

Molar mass of BCl_3 = 117.17 g/mole

Molar mass of B_2H_6 = 27.66 g/mole

Molar mass of LiCl = 42.39 g/mole

First we have to calculate the moles of LiH and BCl_3.

\text{Moles of }LiH=\frac{\text{Mass of }LiH}{\text{Molar mass of }LiH}=\frac{90718.5g}{7.95g/mole}=11411.13moles

\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{453592g}{117.17g/mole}=3871.23moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

6LiH+2BCl_3\rightarrow B_2H_6+6LiCl

From the balanced reaction we conclude that

As, 6 moles of LiH react with 1 mole of BCl_3

So, 11411.13 moles of LiH react with \frac{11411.13}{6}=1901.855 moles of BCl_3

From this we conclude that, BCl_3 is an excess reagent because the given moles are greater than the required moles and LiH is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 3871.23 - 1901.855 = 1969.375 moles

Total excess reactant = 3871.23 moles

Now we have to determine the percent of excess reactant (BCl_3).

\% \text{ excess reactant}=\frac{\text{Moles of remaining excess reactant}}{\text{Moles of total excess reagent}}\times 100

\% \text{ excess reactant}=\frac{1969.375}{3871.23}\times 100=50.87\%

The percent of excess reactant is, 50.87 %

Now we have to calculate the moles of B_2H_6.

As, 6 moles of LiH react to give 1 mole of B_2H_6

So, 11411.13 moles of LiH react to give \frac{11411.13}{6}=1901.855 moles of B_2H_6

Now we have to calculate the mass of B_2H_6.

\text{Mass of }B_2H_6=\text{Moles of }B_2H_6\times \text{Molar mass of }B_2H_6

\text{Mass of }B_2H_6=(1901.855mole)\times (27.66g/mole)=52605.3093g

Now we have to calculate the percent yield of B_2H_6.

\%\text{ yield of }B_2H_6=\frac{\text{Actual yield of }B_2H_6}{\text{Theoretical yield of }B_2H_6}\times 100=\frac{20411.7g}{52605.3093g}\times 100=38.80\%

The percent yield of B_2H_6 or percent conversion of LiH to B_2H_6 is, 38.80 %

Now we have to calculate the moles of LiCl.

As, 6 moles of LiH react to give 6 mole of LiCl

So, 11411.13 moles of LiH react to give 11411.13 moles of LiCl

Now we have to calculate the mass of LiCl.

\text{Mass of }LiCl=\text{Moles of }LiCl\times \text{Molar mass of }LiCl

\text{Mass of }LiCl=(11411.13mole)\times (42.39g/mole)=483717.8007g=1066.42lb

The mass of LiCl produced is, 1066.42 lb

8 0
2 years ago
If a 0.4856 gram sample of khp is dissolved in sufficient water to prepare 250 ml of solution, and 25 ml of the solution require
dangina [55]

Mass of potassium hydrogen pthalate KHP is 0.4856 g, its molar mass is 204.22 g/mol, number of moles of KHP can be calculated as follows:

n=\frac{m}{M}

Here, m is mass and M is molar mass, putting the values,

n=\frac{0.4856 g}{204.22 g/mol}=0.00237 mol

This will be number of moles of NaOH at equivalent point.

Detailed calculations:

Molarity is defined as number of moles in 1 L of solution, for 250 mL of solution, molarity will be:

M=\frac{0.00237 mol}{250 \times 10^{-3}L}=0.009511 M

For 25 mL, apply dilution law as follows:

M_{1}V_{1}=M_{2}V_{2}

Putting the values,

0.009511\times 250=M_{2}\times 25 mL

On rearranging,

M_{2}=\frac{0.009511\times 250}{25}=0.09511 M

Convert molarity into number of moles,

n=M\times V=0.09511 mol/L\times 25\times 10^{-3}L=0.00237 mol

At equivalent point, number of moles of KHP will be equal to NaOH, thus, number of moles of NaOH will be 0.00237 mol.

Calculation for molarity:

Volume of NaOH is 18.75 mL, thus, molarity can be calculated as follows:

M=\frac{n}{V}

Putting the values,

M=\frac{0.00237 mol}{18.75\times 10^{-3}L}=0.1264 M

Therefore, molarity of NaOH is 0.1264 M

7 0
2 years ago
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