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KatRina [158]
3 years ago
15

After an airplane takes off from an airport near sea level, it climbs to a cruising altitude 10,000m above sea level. How does t

he climb affect the weight and mass of the airplane?
Chemistry
1 answer:
juin [17]3 years ago
8 0

Explanation:

As the plane accelerates in its climb towards its cruising altitude, the weight of the plane and its passengers is less as compared to on the earth's surface. The mass, nonetheless, remains the same.

The reason for the change in weight is because as the plane accelerates upwards, it cancels out most of the acceleration by gravity (10m/s²). Gravity applies downwards towards the earth's surface.

Remember that weight is equal to mass * gravity

Reduced gravitational acceleration, will therefore mean a reduction in weight even as mass remains the same.

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What is the mass of 61.9 L of oxygen gas collected at STP?
Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at <em>1 atm, 273 K</em>

<u>Stoichiometry</u>

  • Dimensional Analysis
  • Mole Ratio

Explanation:

<u>Step 1: Define</u>

<em>Identify given.</em>

61.9 L O₂ at STP

<u>Step 2: Convert</u>

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many <em>moles</em> of O₂ is present:

\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from <em>moles</em> to <em>grams</em>:

\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2

Now we deal with sig figs. From the original problem, we are given 3 significant figures. Round your answer to the <u>exact</u> same number of sig figs:

\displaystyle 44.2143 \ \text{g} \ \text{O} \approx \boxed{ 44.2 \ \text{g} \ \text{O}_2 }

∴ our answer is letter choice D.

---

Topic: AP Chemistry

Unit: Stoichiometry

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3 years ago
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A sample of a gas is occupying a 1500 ml container at a pressure of 3.4 atm and a temperature of 25 oc. if the temperature is in
Salsk061 [2.6K]

The new pressure will be 7.65 atm

<h3>General gas law</h3>

The problem is solved using the general gas equation:

P1V1/T1 = P2V2/T2

In this case, P1 = 3.4 atm, V1 = 1500 mL, T1 = 25 ^O C, V2 = 2000 mL, and T2 = 75  ^O C

What we are looking for is P2.

Thus, P2 = P1V1T2/T1V2

          = 3.4 x 1500 x 75/25 x 2000 = 382500/50000 = 7.65 atm

More on general gas laws can be found here: brainly.com/question/2542293

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2 years ago
A 4.0 g sample of iron was heated from 0°C to 20.°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this pie
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Answer:

Explanation:i joule is equal to 0.238902957619 calories so 1251 joules is equal to 298.87 calories divided by 25.0 degrees centigrade is equal to 11.95 calories divided by the 35.2 gram sample weight to get the calories per gram per degree centigrade would come to 0.3396 calories/gram degree centigrade. Presumably this, if correct, could be used to obtain the metal in question by consulting a chart or table with specific heats of various metals because they should always be the same specific heat for each metal.

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School Math. Homework.
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Answer:

The answer is TRUE!!!

Explanation:

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