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3 years ago

Plz help me with this question, I am stuck

Chemistry

1 answer:

zhenek [66]3 years ago

7 0

i only know that E is one of them

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After eating, nutrients must be oved to other parts of the body. Which of the following systems will transport these nutrients.

V125BC [204]

The circulatory system

4 0

3 years ago

Read 2 more answers

If the energy difference between two electronic states is 214.68 kJ / mol , calculate the frequency of light emitted when an ele

atroni [7]

${\qquad\qquad\huge\underline{{\sf Answer}}}$

Here we go ~

Energy difference btween the two electronic states can be expressed as :

${ \qquad \sf \dashrightarrow \: \Delta E = h\nu}$

[ h = planks constant, ${\: \nu }$ = frequency ]

$\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }$

$\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}$

$\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz$

5 0

1 year ago

The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago

Solve

stiv31 [10]

The volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

<h3>How to calculate volume?</h3>

The volume of a given gas can be calculated using the following Charle's law equation:

V1/T1 = V2/T2

Where;

T1 = initial temperature
T2 = final temperature
V1 = initial volume
V2 = final volume

V1 = 9L
V2 = ?
T1 = 325K
T2 = 450K

9/325 = V2/450

325V2 = 4050

V2 = 4050/325

V2 = 12.46L

Therefore, the volume of a gas that occupies 9 L at a temperature of 325K is 12.46L.

Learn more about volume at: brainly.com/question/2817451

7 0

2 years ago

Solve for b if R = x(A + B)

Tatiana [17]

This is what I got. Hope it helps :)

7 0

3 years ago