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14 days ago

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What stress will shift the following equilibrium system to the right? 2SO2(g) + O2(g) ⇌ 2SO3(g); ΔH= –98.8 kJ/mol Decreasing con

centration of SO2 Decreasing temperature Increasing concentration of SO3 Increasing volume

1 answer:

Alekssandra [968]14 days ago

3 0

Answer:

Lowering the temperature.

Explanation:

According to Le Châtelier's principle, when an equilibrium is disturbed by an outside influence, the system reacts by shifting in the direction that counteracts that disturbance and restores balance.
Let's analyze the given options:

1) Reducing the concentration of SO₂,

A decrease in SO₂ concentration causes the reaction to move leftward to compensate for this drop.

2) Lowering the temperature,

Given ΔH = –98.8 kJ/mol, which implies the reaction releases heat and is exothermic.

Dropping the temperature means there's less heat, prompting the equilibrium to shift right to produce more heat and offset the change.

This is the correct option.

3) Increasing SO₃ concentration,

An increase in SO₃ shifts the reaction left to reduce the SO₃ levels.

4) Raising the volume,

Increasing volume lowers the pressure, causing the equilibrium to shift towards the side with more gas moles—that is, the left side.

Therefore, decreasing the temperature is the correct choice.

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You want to prepare a solution with a concentration of 200.0μM from a stock solution with a concentration of 500.0mM. At your di

lorasvet [956]

Answer:

1) This dilution plan will yield a 200μM solution.

2) This dilution plan will not yield a 200μM solution.

3) This dilution plan will not yield a 200μM solution.

4) This dilution plan will yield a 200μM solution.

5) This dilution plan will yield a 200μM solution.

Explanation:

Convert the initial molarity into molar form as shown.

$500mM = 500mM \times (\frac{1M}{1000M})= 0.5M$

Let's examine the following serial dilution processes.

1)

Dilute 5.00 mL of the stock solution to 500 mL. Then take 10.00 mL of this new solution and dilute it further to 250 mL.

Concentration of 500 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{500 mL}= 5 \times 10^{-3}M$

10 mL of this solution is further diluted to 250 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(5 \times 10^{-3}M)(10.0mL)}{250 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

2)

Dilute 5.00 mL of the stock solution to 100 mL. Then take 10.00 mL of this new solution and dilute to 1000 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5.00mL)}{100 mL}= 2.5 \times 10^{-2}M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(2.5 \times 10^{-2}M)(10.0mL)}{1000 mL}= 2.5 \times 10^{-4}M$

Convert μM:

$2.5 \times 10^{-4}M = (2.5 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 250 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

3)

Dilute 10.00 mL of the stock solution to 100 mL, followed by diluting 5 mL of that new solution to 100 mL.

Concentration of 100 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{100 mL}= 0.05M$

5 mL of this solution is diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.05M)(5mL)}{1000 mL}= 0.25 \times 10^{-4}M$

Convert μM:

$0.25 \times 10^{-4}M = (0.25 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 25 \mu M$

Thus, this dilution scheme will not yield a 200μM solution.

4)

Dilute 5 mL of the stock solution to 250 mL. Then take 10 mL of this new solution and further dilute it to 500 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(5mL)}{250 mL}= 0.01M$

10 mL of this solution is further diluted to 500 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.01M)(10mL)}{500 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

5)

Dilute 10 mL of the stock solution to 250 mL. Then take another 10 mL of this new solution and dilute it to 1000 mL.

Concentration of 250 mL solution:

$M_{2}= \frac{M_{1}V_{1}}{V_{2}}= \frac{(0.5M)(10mL)}{250 mL}= 0.02M$

10 mL of this solution is further diluted to 1000 mL

$M_{final}= \frac{M_{2}V_{2}}{V_{final}}= \frac{(0.02M)(10mL)}{1000 mL}= 2 \times 10^{-4}M$

Convert μM:

$2 \times 10^{-4}M = (2 \times 10^{-4}M)(\frac{1 \mu M}{10^{-6}M})= 200 \mu M$

Thus, this dilution scheme will yield a 200μM solution.

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9 days ago

The recommended daily intake of potassium ( K ) is 4.725 g . The average raisin contains 3.513 mg K . Fill in the denominators o

KiRa [971]

Explanation:

It is established that 1 gram is equivalent to 1000 milligrams. We can express this mathematically in the following way.

$\frac{1 g}{1000 mg}$ or $\frac{1000 mg}{1 g}$

Thus, to convert grams to milligrams, we simply multiply the number by 1000. Conversely, for converting milligrams back to grams, we divide by 1000.

4 0

9 days ago

Write the balanced molecular and net ionic equation for the reaction that occurs when the contents of the two beakers are added

alisha [964]

1) reacting hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).

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5 days ago

A 0.680 M Ca(OH)2 solution was prepared by dissolving 55.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol

castortr0y [923]

Convert 55.0g Ca(OH)2 to moles.

The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.

To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.

If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.

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6 days ago

Read 2 more answers

A metallic object holds a charge of −4.8 × 10−6 C. What total number of electrons does this represent? (e = 1.6 × 10−19 C is the

Anarel [852]

Answer:

$n=3.0\times 10^{13}$

Explanation:

Charge of one electron = $-1.6\times 10^{-19}\ C$

The formula for calculating charge is:

$Charge=n\times q_e$

Given that: Charge = $-4.8\times 10^{-6}\ C$

$-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})$

$n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}$

Total electrons, n = $3.0\times 10^{13}$

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13 days ago