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15 days ago

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Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

5.0 grams of O2 with 6.0 grams of S. What is the % yield of SO3 in this experiment? S (s) + O2 (g) → SO3 (g) (not balanced)

1 answer:

VMariaS [1K]15 days ago

6 0

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %

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HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

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Answer:

Quantity of $H_{2}O$ generated will be reduced to fifty percent of its initial amount.

Explanation:

Equilibrium reaction: $HCl+NaOH\rightarrow NaCl+H_{2}O$

In accordance with the balanced equation, 1 mol of HCl interacts with 1 mol of NaOH leading to the formation of 1 mol of $H_{2}O$

<pif the="" quantities="" of="" reactants="" and="" hcl="" are="" diminished="" by="" half="" it="" results="" in:="">

0.5 mol of HCl interacting with 0.5 mol of NaOH yielding 0.5 mol of $H_{2}O$ .

Consequently, it is clear that the total of $H_{2}O$ produced will be halved if the quantities of the reactants are halved.

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14 days ago

In order to use a pipet, place a ____________ at the top of the pipet. Use this object to fill the pipet such that the _________

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Answer:

The right responses are "bulb or pump; meniscus; outside".

Explanation:

Pipets are essential tools in laboratory settings. They are designed for transferring liquids from one vessel to another. First, a bulb or pump is attached to the top to empty the pipet completely. Next, fill the pipet until the meniscus (the curved top of the liquid) aligns with the measurement line corresponding to the volume needed. Finally, dispense the liquid into a second container and make sure to eliminate the last drop beyond the pipet tip.

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8 days ago

The ionic radius of a sodium ion is 2.27 angstroms (A) . What is this length in um

lions [985]

$\boxed{\sf 1Å=10^{-10}m}$

$\\ \rm\longmapsto 2.27Å$

$\\ \rm\longmapsto 2.27\times 10^{-10}m$

$\\ \rm\longmapsto 0.227\times 10^{-9}m$

$\\ \rm\longmapsto 0.0227\times 10^{-8}m$

$\\ \rm\longmapsto 0.00227\times 10^{-7}m$

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12 days ago

Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne

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Answer:

The generation of static electricity occurs when two surfaces are rubbed together. This process causes a transfer of electrons, resulting in a build-up of negative charge. For instance, when you shuffle on a carpet, the friction creates multiple contact points which allow electrons to move onto you, thus accumulating a static charge. Touching another individual or object can lead to a sudden discharge, experienced as an electric shock.

In a similar way, rubbing a balloon against your hair generates opposite static charges on both your hair and the balloon. As you gently pull the balloon away from your head, the attraction between these opposite charges can be observed, causing your hair to rise.

Materials

• Balloon

• Woolen item (like a sweater, scarf, or yarn ball)

• Stopwatch

• Wall

• Partner (optional)

Preparation

• Inflate the balloon and secure the end.

• Have your partner ready to time with the stopwatch.

Procedure

• Grip the balloon with minimal hand coverage, such as holding it with just your thumb and index finger, or by its tied neck.

• Rub the balloon on the wool item once, making sure to go in one direction only.

• Press the rubbed side of the balloon against the wall and let go. Is it adhering to the wall? If it's stuck, your partner should start the stopwatch to measure how long it stays there. If it doesn’t stick, continue to the next step.

• Briefly touch the balloon to a metal object. Why is this step necessary?

• Repeat this procedure, but each time increase the number of rubs against the woolly item, ensuring the direction remains the same (do not rub back and forth).

Observations and results

As you increase the number of times you rub the balloon on the woolly material, does the duration of its adhesion to the wall increase?

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Once the balloon is sufficiently charged negatively by repeated rubbing, it will adhere to the wall. Though the wall typically has a neutral charge, its internal charges can realign such that a positively charged region can attract the negatively charged balloon. Since the wall is also an insulator, the charge does not dissipate instantly. However, when the balloon is in contact with a metal object, the excess electrons from the balloon flow into the metal quickly, making the balloon lose its attraction and peel away.

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4 days ago

Read 2 more answers

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

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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