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Scilla [17]
3 years ago
7

Select the statements that best describe the properties of an intramolecular sn2 reaction mechanism.

Chemistry
1 answer:
ira [324]3 years ago
3 0

Answer:

stereospecific: 100 % Inversion of configuration at the reaction Site

bimolecular at rate-determining step

second order

rate is governed by steric effects

Explanation:

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Answer:

4

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the actual change in phase is 4 on the diagram

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What structure actively moves water through a sponge?
matrenka [14]

Answer:

Answer is flagella

Explanation:

I hope it's helpful!

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Explain the relationship between forward and reverse
serious [3.7K]

If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish equilibrium. If a chemical reaction is at equilibrium and experiences a change in pressure, temperature, or concentration of products or reactants, the equilibrium shifts in the opposite direction to offset the change. This page covers changes to the position of equilibrium due to such changes and discusses briefly why catalysts have no effect on the equilibrium position.

For example, if the system is changed in a way that increases the concentration of one of the reacting species, it must favor the reaction in which that species is consumed. In other words, if there is an increase in products, the reaction quotient, Qc, is increased, making it greater than the equilibrium constant, Kc.

8 0
3 years ago
2-- What is the [H3O+] in a solution with [OH-] = 1 x 10-12 M?<br>​
Arte-miy333 [17]

Answer:

The answer is 0.01 M

Explanation:

The problem is solved by applying the expression for ionic product of water as follows:

Kw = [H₃O⁺] [OH⁻]

Where Kw is the ionic product of water and it is 1.10⁻¹⁴ M at⁴ 25ºC. As [OH⁻]= 1.10⁻¹² M, [H₃O⁺] will be:

[H₃O⁺]= Kw/ [OH⁻]= 1.10⁻¹⁴M/1.10⁻¹²M= 0.01 M

5 0
3 years ago
A solution contains 10.20 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL
AveGali [126]

The question is incomplete, here is the complete question:

A solution contains 10.20 g of unknown compound dissolved in 50.0 mL  of water. (Assume a density of 1.00 g/mL  for water.) The freezing point of the solution is -3.21°C. The mass percent composition of the compound is 60.98% C , 11.94% H , and the rest is O.

What is the molecular formula of the compound?

<u>Answer:</u> The molecular formula for the given organic compound is C_6H_{14}O_2

<u>Explanation:</u>

  • To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of water = 1 g/mL

Volume of water = 50.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{50.0mL}\\\\\text{Mass of water}=(1g/mL\times 50.0mL)=50g

Depression in freezing point is defined as the difference in the freezing point of pure solution and the freezing point of solution

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{freezing point of solution}

  • To calculate the depression in freezing point, we use the equation:

\Delta T_f=i\times K_f\times m

Or,

\text{Freezing point of pure solution}-\text{freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution (water) = 0°C

Freezing point of solution = -3.21°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal boiling point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute = 10.20 g

M_{solute} = Molar mass of solute = ?

W_{solvent} = Mass of solvent (water) = 50.0 g

Putting values in above equation, we get:

(0-(-3.21))^oC=1\times 1.86^oC/m\times \frac{10.20\times 1000}{M_{solute}\times 50}\\\\M_{solute}=\frac{1\times 1.86\times 10.20\times 1000}{3.21\times 50}=118.2g

<u>Calculating the molecular formula:</u>

We are given:

Percentage of C = 60.98 %

Percentage of H = 11.94 %

Percentage of O = (100 - 60.98 - 11.94) % = 27.08 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 60.98 g

Mass of H = 11.94 g

Mass of O = 27.08 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{60.98g}{12g/mole}=5.082moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{11.94g}{1g/mole}=11.94moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.08g}{16g/mole}=1.69moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.69 moles.

For Carbon = \frac{5.082}{1.69}=3

For Hydrogen = \frac{11.94}{1.69}=7.06\approx 7

For Oxygen = \frac{1.69}{1.69}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 7 : 1

The empirical formula for the given compound is C_3H_7O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 118.2 g/mol

Mass of empirical formula = 59 g/mol

Putting values in above equation, we get:

n=\frac{118.2g/mol}{59g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(3\times 2)}H_{(7\times 2)}O_{(1\times 2)}=C_6H_{14}O_2

Hence, the molecular formula for the given organic compound is C_6H_{14}O_2

8 0
3 years ago
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