What bond distance is expected to be longest?1. A carbon-carbon bond with a bond order of 2
2. A carbon-carbon bond with a bond order of 3
3. A carbon-carbon bond with a bond order of O
4. carbon-carbon bond with a bond order of 1​

Answers

Answer 1
Answer:

Answer:

Bond length of C=C is largest(134 pm) because both the carbon atoms have same electronegativity. In case of C=O. and C=N carbon is bonded to highly electronegative atoms so bond length is shoreter as compared to C=C


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To what volume (in mL) would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN?

Answers

Answer:

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Explanation:

Dilution is the reduction of the concentration of a chemical in a solution and consists simply of adding more solvent.

In a dilution the amount of solute does not vary. But as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

In a solution it is fulfilled:

Ci* Vi = Cf* Vf

where:

  • Ci: initial concentration
  • Vi: initial volume
  • Cf: final concentration
  • Vf: final volume

In this case:

  • Ci= 1.40 M
  • Vi= 20 mL
  • Cf= 0.088 M
  • Vf= ?

Replacing:

1.40 M* 20 mL= 0.088 M* Vf

Solving:

Vf=(1.40 M* 20 mL)/(0.088 M)

Vf= 318.18 mL

To 318.18 mL would you need to dilute 20.0 mL of a 1.40 M solution of LiCN to make a 0.0880 M solution of LiCN

Final answer:

To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.

Explanation:

To dilute a solution, you can use the formula:

M1V1 = M2V2

where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume. Rearranging the formula, we can solve for V2:

V2 = (M1 · V1) / M2

Plugging in the values given:

V2 = (1.40 M · 20.0 mL) / 0.0880 M = 318.18 mL

To make a 0.0880 M solution of LiCN, you would need to dilute 20.0 mL of the 1.40 M solution to a final volume of 318.18 mL.

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By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution. A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 13.0g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution Molarity= O.822 M

Answers

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = V_1=11.00 mL

Concentration of stock solution = M_1=0.823 M

Volume of stock solution after dilution = V_2=50.00 mL

Concentration of stock solution after dilution = M_2=?

M_1V_1=M_2V_2 ( dilution )

M_2=(0.823 M* 11.00 mL)/(50 ,00 mL)=0.18106 M

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}* Volume (L)}

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

Molarity=(13.0 g)/(158 g/mol* 0.100 L)=0.823 mol/L

0.823 Molar is the molarity of the solution.

Final answer:

To determine the molarity of the resulting solution, we can use the formula M1V1 = M2V2. Plugging in the given values, we find that the molarity of the resulting solution is 0.180 MM.

Explanation:

To determine the molarity of the resulting solution, we need to use the formula:

M1V1 = M2V2

Where M1 is the molarity of the stock solution, V1 is the volume of the stock solution used, M2 is the molarity of the resulting solution, and V2 is the final volume of the resulting solution.

Using the given values, we have:

M1 = 0.823 MM

V1 = 11.00 mL

V2 = 50.00 mL

Substituting these values into the formula, we can find the molarity of the resulting solution.

M2 = (M1 * V1) / V2

Plugging in the values:

M2 = (0.823 MM * 11.00 mL) / 50.00 mL = 0.180 MM

The molarity of the resulting solution is 0.180 MM.

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An imaginary line dividing the earth's surface into two hemisphere the northern and southern hemisphere, it is locatedat 0⁰, which of the following imaginary lines is being described? a.equator b.latitude c.longitude d.prime meridian​

Answers

Answer:

A. Equator

Explanation:

The equator is located in the centre of the Earth, dividing the northern and southern hemispheres.

Answer:

Equator because equator divides the earth into Northern and Southern hemisphere.

7. A change of state is a(n)_process.
a. Irreversible
b. Reversible

Answers

Answer: b

A change of state is a reversible process.

B that’s the answer

Good luck

Match the vocabulary word with the defination.___1.Hypothesis
___2.Variables
___3.Conclusion
___4.Scientific Method
___5.Procedure

a.The steps you take to complete the experiment
b.Factors that changes in an experiment
c.A possible solution to a problem
d.The result of the experiment
e.The process scientist follow to complete an investigation

Answers

Hypothesis - A possible solution to a problem

Variables - Factors that changes in an experiment

Conclusion - The result of the experiment

Scientific Method - The process scientists follow to complete an investigation

Procedure - The steps you take to complete the experiment

Answer:

1. d

2. b

3. d

4. e

5. a

explanation:

there's nothing else to explain

A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of the solution. The chemical reaction that takes place is represented by the following equation. Pb(NO3)2(s) + NaI(aq) → PbI2(s) + NaNO3(aq) How many moles of PbI2 can be produced?/Identify the limiting reactant/If the actual yield was 0.550 g of PbI2, what is the percent yield?

Answers

Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

Further explanation

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%