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3 years ago

A system is in equilibrium when the rate of the forward reaction is _____ the rate of the reverse reaction.

Chemistry

1 answer:

NemiM [27]3 years ago

3 0

Answer:

Option 4) equal to

Explanation

When having a reversible reaction, this reaction (or system) reaches the balnace, it is observed that relative quantitives for all compounds (this is, reactives and products), remain constant

The species concentration does not change over time, and; in the same way; there are no physical changes as time goes by

For example, for the following reaction :

aA + bB ⇔ cC + dD

Compounds A and B react to give products C and D. The two way arrows means that this systme has reached the balance (is in equilibrium)

You might be interested in

How much mass does 1 mol of O2 gas have?

gregori [183]

Answer:

Option A.

Explanation:

1 mol of anything contains 6.02×10²³ particles.

We know that 1 mol of oxygen gas contains 2 moles of O.

1 mol of oxygen weighs 16 g/mol, the mass for 1 molecule of O.

By the way, the mass for 1 mol of O₂ may be:

Option A → 16 g/mol . 2 mol

32 g

Oyxgen is a dyatomic molecule, that's why we have 2 moles of O.

Another example can be:

1 mol of water (H₂O) contains 2 moles of H and 1 mol of O.

4 0

3 years ago

A gas sealed in a drum container occupies 7.70 L at a sizzling 325C degrees. What will the volume be if the drum is quickly chi

grigory [225]

The volume of the drum after chilled quickly is 3.6 L.

<u>Explanation:</u>

As per the Charles Law, for any gas at constant pressure, volume of the gas is in direct proportion with the temperature measured in Kelvin. As the temperature increases, the gas expands and vice versa.

$$\frac{V1}{T1} = \frac{V2}{T2} \\$

After chilling the drum, the volume will get reduced and it can be found as,

V2 = $$\frac{V1T2}{T1}$

= $$\frac{7.7 L \times 282 K}{598 K}$

= 3.6 L

Here the drum is chilled quickly, that is temperature is reduced and so the volume also decreases.

3 0

3 years ago

A large balloon is initially filled to a volume of 25.0 L at 353 K and a pressure of 2575 mm Hg. What volume of gas will the bal

bija089 [108]

Answer:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

Given ,

V₁ = 25.0 L

V₂ = ?

P₁ = 2575 mm Hg

Also, P (atm) = P (mm Hg) / 760

P₁ = 2575 / 760 atm = 3.39 atm

P₂ = 1.35 atm

T₁ = 353 K

T₂ = 253 K

Using above equation as:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}$

$\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}$

Solving for V₂ , we get:

45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.

4 0

4 years ago

Calculate the pH of the solution formed when 45.0 mL of 0.100M NaOH solution is added to 50.0 mL of 0.100M CH3COOH (Ka for aceti

pav-90 [236]

Answer:

pH of soltion will be 5.69

Explanation:

The pH of the solution will be due to excessive acid left and the salt formed. Thus, it will form a buffer solution.

The pH of buffer solution is calculated from Henderson Hassalbalch's equation, which is:

$pH=pKa+log(\frac{[salt]}{[acid]} )$

$pKa=-log(Ka)$

$pKa=-log(1.8X10^{-5})=4.74$

The moles of acid taken :

$moles=molarityXvolume=0.1X50=5mmol$

The moles of base taken:

$moles=molarityXvolume=0.1X45=4.5mmol$

The moles of acid left after reaction :

$5-4.5=05mmol$

The moles of salt formed = 4.5mmol

Putting values in equation

$pH=pKa+log(\frac{[salt]}{[acid]} )=4.74+log(\frac{4.5}{0.5})=5.69$

8 0

3 years ago

Complete the equation determine if there are precipitates write the balanced ionic and net ionic equations. If there is no preci

alukav5142 [94]

You may find bellow the balanced chemical equations.

Explanation:

Molecular equations:

3 Sr(NO₃)₂ (aq) + 2 K₃PO₄ (aq) → Sr₃(PO₄)₂ (s) + 6 KNO₃ (aq)

2 NaOH (aq) + Ni(NO₃)₂ (aq) = Ni(OH)₂ (s) + 2 NaNO₃ (aq)

Ionic equations:

3 Sr²⁺ (aq) + 6 NO₃⁻ (aq) + 6 K⁺ (aq) + 2 PO₄³⁻ (aq) → Sr₃(PO₄)₂ (s) + 6 K⁺ (aq) + 6 NO₃⁻ (aq)

2 Na⁺ (aq) + 2 OH⁻ (aq) + Ni²⁺ (aq) + 2 NO₃⁻ (aq) = Ni(OH)₂ (s) + 2 Na⁺ (aq) + 2 NO₃⁻ (aq)

To get the net ionic equation we remove the spectator ions:

3 Sr²⁺ (aq) + 2 PO₄³⁻ (aq) → Sr₃(PO₄)₂ (s)

2 OH⁻ (aq) + Ni²⁺ (aq) = Ni(OH)₂ (s)

Learn more about:

net ionic equations

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4 0

3 years ago