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Mrrafil [7]
2 years ago
11

Enter your answer in the provided box. An industrial chemist introduces 2.0 atm of H2 and 2.0 atm of CO2 into a 1.00−L container

at 25.0°C and then raises the temperature to 700°C, at which Kc = 0.534: H2(g) + CO2(g) ⇌ H2O (g) + CO(g) How many grams of H2 are present at equilibrium?
Chemistry
1 answer:
PSYCHO15rus [73]2 years ago
8 0

Answer: 0.0944 gram of H2

Explanation:

Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:

2.0 atm x (973 K / 298 K) = 6.53 atm

Where

Kc = Kp because the moles of product equals the moles of reactants.

At equilibriuim, the amounts are

P(H2) = 6.53 - x

P(CO2) = 6.53 - x

P(H2O) = x

P(CO) = x

Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]

Take the square root of each side

(.534)^0.5 = x / (6.53 - x)

x = 0.731 (6.53 - x)

x = 4.77 - 0.731x

1.731x = 4.77

x = 4.77 / 1.731 = 2.76 atm

P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm

P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm

PV = nRT

n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2

0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g

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Calculate the molarity of sodium chloride in a half-normal saline solution (0.45% NaCl). The molar mass of NaCl is
Rus_ich [418]

Answer:

0.077 M

Explanation:

Data Given :

The concentration of half normal (NaCl) saline = 0.45g / 100 g

So,

Volume of Solution = 100 g = 100 mL

Volume of Solution in Liter = 100 mL / 1000

Volume of Solution = 0.1 L

molar mass of NaCl = 58.44 g/mol

Molarity:

Molarity is the representation of the solution. It is amount of solute in moles per liter of solution and represented by M

Formula used for Molarity

                M = moles of solute / Liter of solution . . . . . . . . . . (1)

Now to find number of moles of Nacl

                no. of moles of NaCl = mass of NaCl / molar mass

                no. of moles of NaCl = 0.45g / 58.44 g/mol

               no. of moles of NaCl = 0.0077 g

Put values in the eq (1)

                  M = moles of solute / Liter of solution . . . . . . . . . . (1)

                  M = 0.0077 g / 0.1 L

                  M = 0.077 M

So the molarity of half-normal saline solution (0.45% NaCl) = 0.077 M

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2 years ago
Explain why you hear a “whoosh” sound when you open a can containing a carbonated drink. Which gas law applies?
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Carbonated drinks have the air under pressure so that carbon bubbles are forced into the drink, keeping it carbonated. So when you open a can, the air under pressure in the can comes out of the can at a high speed, making a "whooshing" sound. The gas law that applies to this concept is the Boyle's Law (PV=k or P1V1=P2V2).

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2 years ago
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Lipids cannot be compressed since there is only a small distance between the molecules when bonded
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2 years 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:

When one object is rubbed against another, static electricity can be created. This is because the rubbing creates a negative charge that is carried by electrons. The electrons can build up to produce static electricity. For example, when you shuffle your feet across a carpet, you are creating many surface contacts between your feet and the carpet, allowing electrons to transfer to you, thereby building up a static charge on your skin. When you touch another person or an object, you can suddenly discharge the static as an electrical shock.

Similarly, when you rub a balloon on your head it causes opposite static charges to build up both on your hair and the balloon. Consequently, when you pull the balloon slowly away from your head, you can see these two opposite static charges attracting one another and making your hair stand up.

Materials

• Balloon

• An object made out of wool (such as a sweater, scarf, blanket or ball of yarn)

• Stopwatch

• A wall

• A partner (optional)

Preparation

• Blow up the balloon and tie off the end.

• Have your partner prepare to use the stopwatch.

Procedure

• Hold the balloon in a way that your hand covers as little of its surface area as possible, such as by using only your thumb and pointer finger or by gripping the balloon by its neck where it is tied off.

• Rub the balloon on the woolly object once, in one direction.

• Hold the balloon up on the wall with the side that was rubbed against the wool facing the wall, then release it. Does the balloon stay stuck on the wall? If the balloon stays stuck, have your partner immediately start the stopwatch to time how long the balloon remains bound to the wall. If the balloon does not stick, move to the next step.

• Touch the balloon to a metal object. Why do you think this is important to do?

• Repeat the above process but each time increase the number of times you rub the balloon on the woolly object. Rub the balloon in the same direction each time. (Do not rub the balloon back and forth.)

Observations and results

In general, did the balloon stick to the wall for a longer amount of time as you increased the number of times you rubbed the balloon on the woolly object?

Wool is a conductive material, which means it readily gives away its electrons. Consequently, when you rub a balloon on wool, this causes the electrons to move from the wool to the balloon's surface. The rubbed part of the balloon now has a negative charge. Objects made of rubber, such as the balloon, are electrical insulators, meaning that they resist electric charges flowing through them. This is why only part of the balloon may have a negative charge (where the wool rubbed it) and the rest may remain neutral.

When the balloon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall. Although the wall should normally have a neutral charge, the charges within it can rearrange so that a positively charged area attracts the negatively charged balloon. Because the wall is also an electrical insulator, the charge is not immediately discharged. However, because metal is an electrical conductor, when you rub the balloon against metal the extra electrons in the balloon quickly leave the balloon and move into the metal so the balloon is no longer attracted and does not adhere.

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2 years ago
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Answer:

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Explanation:

The half-life of K-40 (1.3 billion years) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

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<u>half-lives</u>      <u>    t/yr      </u>       <u>Remaining</u>  

      0               0                          1

      1                1.3  billion              ½

      2              2.6                          ¼

      3              3.9                          ⅛

We see that after 2 half-lives, ¼ of the original mass remains.

Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.

Original  mass = 4 × 2.10 g = \boxed{ \text{8.40 g}}

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