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2 years ago

At equilibrium, no chemical reactions take place. true or false

1 answer:

6 0

False. At equilibrium, the rate of forward reaction is equal to the rate of backward reaction. The net concentration of both products and reactants won't change, but the reactions still take place.

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If the copper cube had a mass of 28.7 grams and a volume of 3.2 mL, what would its density be?

Otrada [13]

Answer:

ρ=m÷v

=28.7÷3.2

≈8.97 g/m³

3 0

3 years ago

For the following reaction, find the value of Q and predict the direction of change, given that a 1L flask initially contains 2

Tresset [83]

Answer:

C) Q < K, reaction will make more products

Explanation:

1/8 S8(s) + 3 F2(g) ↔ SF6(g)

∴ Kc = 0.425 = [ SF6 ] / [ F2 ]³

∴ Q = [ SF6 ] / [ F2 ]³

∴ [ SF6 ] = 2 mol/L

∴ [ F2 ] = 2 mol/L

⇒ Q = ( 2 ) / ( 2³)

⇒ Q = 0.25

⇒ Q < K, reaction will make more products

5 0

2 years ago

What is the chemical process of breaking down food and realeasing energy?

Sergeu [11.5K]

This process is called aerobic respiration.

5 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

To a 25.00 mL volumetric flask, a lab technician adds a 0.150 g sample of a weak monoprotic acid, HA , and dilutes to the mark w

Elis [28]

<u>Answer:</u> The number of moles of weak acid is $4.24\times 10^{-3}$ moles.

<u>Explanation:</u>

To calculate the moles of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}\text{Volume of solution (in L)}}$

We are given:

Volume of solution = 43.81 mL = 0.04381 L (Conversion factor: 1L = 1000 mL)

Molarity of the solution = 0.0969 moles/ L

Putting values in above equation, we get:

$0.0969mol/L=\frac{\text{Moles of KOH}}{0.04381}\\\\\text{Moles of KOH}=4.24\times 10^{-3}mol$

The chemical reaction of weak monoprotic acid and KOH follows the equation:

$HA+KOH\rightarrow KA+H_2O$

By Stoichiometry of the reaction:

1 mole of KOH reacts with 1 mole of weak monoprotic acid.

So, $4.24\times 10^{-3}mol$ of KOH will react with = $\frac{1}{1}\times 4.24\times 10^{-3}=4.24\times 10^{-3}mol$ of weak monoprotic acid.

Hence, the number of moles of weak acid is $4.24\times 10^{-3}$ moles.

6 0

3 years ago