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3 years ago

14

Aqueous acetic acid is neutralized by aqueous potassium hydroxide. True or False

2 answers:

Stels [109]3 years ago

8 0

Answer:

i think its true

Explanation:

liberstina [14]3 years ago

4 0

Answer:

i think its true but I’m not sure

Explanation:

I know that they can. Be mixed

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Which of the following expressions best represents a second order rate law?

Ksivusya [100]

I think the correct answer would be C. The expression that would best represent a second order rate law would be r =k[X][Y]. Reaction with this rate law are those that depend on the concentration of two first order reactants or a second order reactant.

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3 years ago

Can someone help me with this I've been stuck on it for a few days

Fiesta28 [93]

So I haven’t got time to answer all of it for you but the id you look at the picture of the periodic table I’ve added the top number in the red boxes are the groups and the period is how many elements down from the top it is (remember that the hydrogen and helium make up period ONE) so remember to include them when counting the elements as you go down the table

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2 years ago

A chemist prepares a solution of sodium thiosulfate by measuring out of sodium thiosulfate into a volumetric flask and filling t

marshall27 [118]

Mass of sodium thiosulfate $Na_{2}S_{2}O_{3}$ is 110. g

Volume of the solution is 350. mL

Calculating the moles of sodium thiosulfate:

$110. g Na_{2}S_{2}O_{3} * \frac{1 mol Na_{2}S_{2}O_{3}}{158.1 g Na_{2}S_{2}O_{3}}$ = 0.696 mol $Na_{2}S_{2}O_{3}$

Converting the volume of solution to L:

$350. mL * \frac{1 L}{1000 mL} = 0.350 L$

Finding out the concentration of solution in molarity:

$\frac{0.696 mol}{0.350 L} = 1.99 mol/L$

4 0

3 years ago

The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c

crimeas [40]

Answer : The rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $785.0K$ = ?

$Ea$ = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $785.0K$

Now put all the given values in this formula, we get:

$\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]$

$K_2=1.45\times 10^{-2}s^{-1}$

Therefore, the rate constant at 785.0 K is, $1.45\times 10^{-2}s^{-1}$

7 0

3 years ago

11. The back part of the cerebrum deals with hearing<br><br> True<br><br> False

Sunny_sXe [5.5K]

Answer:

The answer is: True

5 0

3 years ago