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mihalych1998 [28]
3 years ago
11

In order to wash out the impurity, you looked up some sketchy method from 1952 which Hank Pym wrote during his Ph. D. under E. J

. Corey. The method requires you add HCl to the mixture. After you rid your sample of any impurities, you take some pH paper and find out your solution has a pH of 3.0. Assuming the pKa of Pym Particles is 4.5 and that the molar absorptivity of the protonated Pym Particle is 4.82 * 102 L/mol*cm and 8.54 * 103 L/mol*cm for 400 nm and 690 nm respectively, calculate the expected A400 and A690 of the solution.

Chemistry
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

516.24

Explanation:

The solved solution is in the attached document.

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Answer:

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Explanation:

5 0
3 years ago
Does carbon ever leave the EARTH? Explain.
GrogVix [38]

Answer:

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Hope this helps!

8 0
3 years ago
Consider the following reaction: CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) The initial rate of the reaction is measured at several di
Ira Lisetskai [31]

Answer : The correct rate law for the reaction is,

\text{Rate}=k[CHCl_3][Cl_2]^{1/2}

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

CHCl_3(g)+Cl_2(g)\rightarrow CCl_2(g)+HCl(g)

Rate law expression for the reaction:

\text{Rate}=k[CHCl_3]^a[Cl_2]^b

where,

a = order with respect to CHCl_3

b = order with respect to Cl_2

Expression for rate law for first observation:

0.0035=k(0.010)^a(0.010)^b ....(1)

Expression for rate law for second observation:

0.0069=k(0.020)^a(0.010)^b ....(2)

Expression for rate law for third observation:

0.0098=k(0.020)^a(0.020)^b ....(3)

Expression for rate law for fourth observation:

0.027=k(0.040)^a(0.040)^b ....(4)

Dividing 1 from 2, we get:

\frac{0.0069}{0.0035}=\frac{k(0.020)^a(0.010)^b}{k(0.010)^a(0.010)^b}\\\\2=2^a\\a=1

Dividing 2 from 3, we get:

\frac{0.0098}{0.0069}=\frac{k(0.020)^a(0.020)^b}{k(0.020)^a(0.010)^b}\\\\1.42=2^b\\b=\frac{1}{2}

Calculation used :

1.42=2^b\\\log (1.42)=b\log 2\\\log (\frac{1.42}{2})=b\\b=0.5=\frac{1}{2}

Thus, the rate law becomes:

\text{Rate}=k[CHCl_3]^1[Cl_2]^{1/2}

7 0
3 years ago
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LuckyWell [14K]

Answer:

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Explanation:

Cost of electricity = $ 0.18 per KW-H

Time = 5 hour per day

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(a) Cost of electricity for 100 W power bulb

0.1 × 5 × 365 × 0.18 =  $ 32.85

(b) Cost of electricity for 0.025 W  fluorescent  bulb

0.025 × 5 × 365 × 0.18 = $ 8.2125

Therefore the cost of electricity  for 100 W power bulb = $ 32.85

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