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77julia77 [94]
3 years ago
10

The book procedure calls for 2,5-dimethylhexane-2,5-diol and concentrated hydrochloric acid to be shaken in a reaction vial. Whi

ch of the following is a hazard associated with this procedure?
A) Hydrochloric acid can cause skin and eye burns, so gloves and goggles must be worn.
B) Hydrochloric acid has irritating vapors, and must be used in the fume hood
C) The vial may build up pressure. Be sure to vent the vial frequently, and keep the glass sash of the fume hood between you and the vial.
D) The vial could be dropped, splashing hydrochloric acid. Keep the vial well inside the hood.
E) All of the above.
Chemistry
1 answer:
san4es73 [151]3 years ago
7 0

Answer:

E) All of the above.

Explanation:

Hello,

Since the acidic nature of the HCl implies its corrosiveness, when it is in contact with the skin and eyes the burning starts immediately, so gloves and goggles must be worn. Next, the fuming hydrochloric acid (37% by mass) is volatile so it gives off even when dissolved into water, so it must be used in the fume hood. Then, since vapors are produced during the chemical reaction, an overpressure could be attained, that's why we must keep the glass sash of the fume hood between us and the vial. As a common risk, the vial could be dropped causing the hydrochloric acid to splash, so we must keep the vial well inside the hood.

Best regards.

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The volume of a gas is 27.5 mL at 22.0°C and 0.974 atm. What will the volume be at 15.0°C and 0.993 atm? Use Ideal Gas Law (PV =
Scorpion4ik [409]

Answer:

26.3 mL

Explanation:

Step 1:

Obtaining an appropriate gas law from the ideal gas equation.

This is illustrated below:

From the ideal gas equation:

PV = nRT

Divide both side by T

PV/T = nR

At this stage, we'll assume the number of mole (n) to be constant.

Note: R is the gas constant.

PV/T = constant.

We can thus, write the above equation as:

P1V1/T1 = P2V2/T2

The above equation is called the general gas equation.

Step 2:

Data obtained from the question. This includes the following:

Initial volume (V1) = 27.5 mL

Initial temperature (T1) = 22.0°C = 22.0°C + 273 = 295K

Initial pressure (P1) = 0.974 atm.

Final temperature (T2) = 15.0°C = 15.0°C + 273 = 288K

Final pressure (P2) = 0.993 atm

Final volume (V2) =..?

Step 3:

Determination of the final volume of the gas using the general gas equation obtained. This is illustrated below:

P1V1 /T1 = P2V2/T2

0.974 x 27.5/295 = 0.993 x V2/288

Cross multiply to express in linear.

295x0.993xV2 = 0.974x27.5x288

Divide both side by 295 x 0.993

V2 = (0.974x27.5x288)/(295x0.993)

V2 = 26.3 mL

Therefore, the new volume of the gas is 26.3 mL

4 0
3 years ago
A hydrogen filled balloon has a bolume of 8.3 L at 36 C and 751 torr. How many moles of hydrogen are inside the balloon?
alina1380 [7]

Answer:

0.33 mol

Explanation:

Given data:

Volume of balloon = 8.3 L

Temperature = 36°C

Pressure = 751 torr

Number of moles of hydrogen = ?

Solution:

Temperature = 36°C (27 +273 = 300 K)

Pressure = 751 torr (751/760= 0.988 atm)

Formula:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

PV  = nRT

0.988 atm × 8.3 L  = n × 0.0821 atm.L/ mol.K  × 300 K

8.2 atm.L = n × 24.63 atm.L/ mol

n = 8.2 atm.L / 24.63 atm.L/ mol

n = 0.33 mol

8 0
3 years ago
What is the formula for flow rate?
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Answer:Solve for X+ 2x=

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3 years ago
Cindy made tea. She started with 300 grams of water at 20 degrees Celsius. She transferred 18,000 calories to the water. What wa
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Answer:

T final = 80°C

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∴ T1 = 20°C = 293 K

∴ T2 = ?

⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)

⇒ (18000 cal)/(300 cal/K) = T2 - 293 K

⇒ T2 = 293 K + 60 K

⇒ T2 = 353 K (80°C)

8 0
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Given:
bulgar [2K]

Answer:

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Explanation:

Given;

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From the combustion reaction above, it can be observed that;

1 mole of methane (CH₄) released 890 kilojoules of energy.

Now, we convert 59.7 grams of methane to moles

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= 3.73125 moles x  -890 kJ/mol

= -3320.81 kJ

Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy

5 0
3 years ago
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