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3 years ago

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Select the statement that is true regarding scientific theories

1 answer:

solniwko [45]3 years ago

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Give the result of the following expression with the correct number of significant figures

Marianna [84]

84 m/s x 31.221 s = 2600 m
-rounded from 2622.564 m because 84 m/s has only two significant figures

4 0

3 years ago

Can anyone answer this question?

-Dominant- [34]

Hello! Those are called the isotopes of those elements!! They have more than just one form and each has it’s own AMU’s so the multiple numbers above different elements, represent the isotopes of those elements. Carbon is a good one just for one example.

Hope this helps you out! Questions please just ask! Thanks!

8 0

3 years ago

A student has two solid blocks that have different masses. The student knows that each block is composed of a pure substance, bu

solniwko [45]

À because they getting block for water not getting in

8 0

3 years ago

Read 2 more answers

A reaction has a rate constant of 2.08 × 10−4 s−1 at 26 oC and 0.394 s−1 at 79 oC . Determine the activation barrier for the rea

leva [86]

<u>Answer:</u> The activation energy of the reaction is 124.6 kJ/mol

<u>Explanation:</u>

To calculate activation energy of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{79^oC}$ = equilibrium constant at 79°C = $0.394s^{-1}$

$K_{26^oC}$ = equilibrium constant at 26°C = $2.08\times 10^{-4}s^{-1}$

$E_a$ = Activation energy of the reaction = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $26^oC=[26+273]K=299K$

$T_2$ = final temperature = $79^oC=[79+273]K=352K$

Putting values in above equation, we get:

$\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol$

Hence, the activation energy of the reaction is 124.6 kJ/mol

3 0

3 years ago

The next 11 questions are related to the titration of 40.00 mL of a 0.0900 M acetic acid solution with 0.0700 M KOH. Assume that

liberstina [14]

Answer:

2.9 is the initial pH of the analyte solution.

Explanation:

The dissociation constant of acetic acid as per theoretical value = $K_a$

$K_a=10^{-4.756}=1.8\times 10^{-5}$

The initial concentration of acetic acid = c = 0.0900 M

$HAc\rightleftharpoons Ac^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[Ac^-][H^+]}{[HAc]}$

$1.8\times 10^{-5}=\frac{x\times x}{(c-x)}$

$1.8\times 10^{-5}=\frac{x\times x}{(0.0900-x)}$

Solving for x:

x = 0.001264 M

$[H^+]=0.001264 M$

The pH of the solution :

$pH=-\log[0.001264]=2.898\approx 2.9$

2.9 is the initial pH of the analyte solution.

4 0

3 years ago