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3 years ago

Need help asap - There is an error in the rate-determining step of the following proposed mechanism.

Chemistry

1 answer:

sdas [7]3 years ago

5 0

Answer:

CH4(g) + Cl2(g) → CH3(g) + HCl(g) and the rate rule is k1 = [CH4][Cl2]

Explanation:

Firstly, we must remember that the substitution reaction between halogens and alkanes to yield halogenoalkanes does not proceed by ionic mechanism rather it proceeds by free radical mechanism.

Now, if we look at the rate determining step as shown in the question, the elementary reaction equation of that step is not balanced.

Hence, the correct elementary reaction equation for the rate determining step and the rate law is; CH4(g) + Cl2(g) → CH3(g) + HCl(g) and the rate rule is k1 = [CH4][Cl2]

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Which natural resource is nonrenewable?

insens350 [35]

Answer: The answer is D, Coal

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3 years ago

A molecule is known to have a molar absorptivity of 18,650 at a certain wavelength. A spectrometer is tuned to this wavelength a

iren2701 [21]

Answer:

17,932.69 g/mol is the molecular weight of the substance.

Explanation:

Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution = 1.04

c = concentration of solution =?

l = length of the cell = 1 cm

$\epsilon$ = molar absorptivity of this solution = 18,650 $M^{-1} cm^{-1}$

Now put all the given values in the above formula, we get the molar absorptivity of this solution.

$1.04=18,650 M^{-1} cm^{-1}\times c)\times (1cm)$

c = $5.576\times 10^{-5} M$

$Concentration=\frac{\text{Mass of compound}}{\text{Molecular mass of compound}\times V}$

V = Volume of the solution in L

Molecular weight of the substance = x

V = 100 mL = 0.1 L

Mass of the substance = 100 mg = 0.1 g

$5.576\times 10^{-5} M=\frac{0.1 g}{x\times 0.1 L}$

x = 17,932.69 g/mol

17,932.69 g/mol is the molecular weight of the substance.

6 0

3 years ago

Once the concentration of the common ion is increased: Select the correct answer below: the solubility of the ionic solid increa

Vaselesa [24]

Answer:

the solubility of the ionic solid decreases

Explanation:

If a salt MX is added to an aqueous solution containing the solute AX, the X^- ion is common to both of the salts. The presence of X^- in the solution will suppress the dissolution of AX compared to the solubility of AX in pure water. This observation is known as common ion effect in chemistry.

The origin of common ion effect is based on Le Chatelier's principle. The addition of a solute will drive the equilibrium position towards the left hand side.

3 0

3 years ago

A chemist makes of nickel(II) chloride working solution by adding distilled water to of a stock solution of nickel(II) chloride

Lady_Fox [76]

Answer:

0.0900 mol/L

Explanation:

<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>

Step 1: Given data

Initial concentration (C₁): 0.135 mol/L

Initial volume (V₁): 220. mL

Final concentration (C₂): ?

Final volume (V₂): 330. mL

Step 2: Calculate the concentration of the final solution

We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂

C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L

3 0

3 years ago

A science teacher has a supply of 20% saline solution and a supply of 50% saline solution. How much of each solution should the

lilavasa [31]

Answer:

Explanation:

$\text{Let}\\\\x-\text{a volume of}\ 20\%\ \text{saline solution}\\\\y-\text{a volume of }\ 50\%\ \text{saline solution}\\\\p\%=\dfrac{p}{100}\\\\20\%=\dfrac{20}{100}=\dfrac{2}{10}=0.2\\\\50\%=\dfrac{50}{100}=\dfrac{5}{10}=0.5\\\\0.2x-\text{a volume of salt in }\ 20\%\ \text{saline solution}\\0.5y-\text{a volume of salt in }\ 50\%\ \text{saline solution}\\\\60mL-\text{a volume of}\ 28\%\ \text{saline solution}\\\\28\%=\dfrac{28}{100}=0.28$

$0.28\cdot60mL=16.8mL-\text{a volume of salt in }\ 28\%\ \text{saline solution}\\\\\text{Equations}\\\\(1)\qquad x+y=60\\(2)\qquad0.2x+0.5y=16.8$

$\left\{\begin{array}{ccc}x+y=60\\0.2x+0.5y=168&\text{multiply both sides by 10}\end{array}\right\\\left\{\begin{array}{ccc}x+y=60&\text{multiply both sides by (-2)}\\2x+5y=168}\end{array}\right\\\underline{+\left\{\begin{array}{ccc}-2x-2y=-120\\2x+5y=168}\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad3y=48\qquad\text{divide both sides by 3}\\.\qquad y=16$

$\text{Put the value of}\ y\ \text{to the first equation:}\\\\x+16=60\qquad\text{subtract 16 from both sides}\\x=44$

6 0

4 years ago