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VLD [36.1K]
3 years ago
10

Two common materials we encounter in our daily lives are radiator coolant in automobiles and dry ice. The coolant in an automobi

le radiator, a solution of antifreeze (a glycol such as ethylene or propylene glycol) and water, freezes at 39°C and boils at 110°C. The temperature of the dry ice (solid carbon dioxide) used in some ice cream vending carts is –78°C. Convert the boiling point of radiator coolant (110°C) to degrees Fahrenheit. Convert the temperature of dry ice (–78°C) to the Kelvin scale.
Chemistry
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

See explanation

Explanation:

Using the formula

°C = (F-32) × 5/9

Where;

°C = temperature in degrees centigrade

F= temperature in Fahrenheit

F= (9/5 ×°C) +32

F= (9/5 × 110) + 32

F= 230°F

To convert -78°C to Kelvin

-78°C + 273 = 195 K

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An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog
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Answer:

1 mole of platinum

Explanation:

To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.

The empirical formula for the compound can be obtained as follow:

Platinum (Pt) = 117.4 g

Carbon (C) = 28.91 g

Nitrogen (N) = 33.71 g

Divide by their molar mass

Pt = 117.4 / 195 = 0.602

C = 28.91 / 12 = 2.409

N = 33.71 / 14 = 2.408

Divide by the smallest

Pt = 0.602 / 0.602 = 1

C = 2.409 / 0.602 = 4

N = 2.408 / 0.602 = 4

The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.

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After studying chemical reactions in school, Jen examined some of the different ways people use synthetic materials. She found t
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Given : 4.50 molte
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You would get 35.................
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Ions are atoms that have gained or lost which subatomic particles?
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A ball is equipped with a speedometer and launched straight upward. The speedometer reading four seconds after launch is shown a
Andrew [12]

Answer:

Question 1: <u>1 s after the motion starts</u>

Question 2: <u>0 (just when the motion starts)</u>

Explanation:

You will need to work with approximates values because the precision of the speedometers is low and you are requested to find approximate times.

<u>1. From the speedometer shown at the right.</u>

You can obtain how long the ball has been falling from the highest altitute it reached using the speed of 10 m/s shown by the speedometer at the right.

  • Free fall equation: Vf = Vo - gt

  • Vo = 0 ⇒ Vf = gt ⇒ t = Vf / g

For this problem, I recommend to work with a rough estimate of g: g = 10 m/s² ( I will tell you why soon)/

  • t = [10 m/s] / [10 m/s²] = 1 s

That is the time falling. Since four seconds after launch have elapsed, the upward time was 3 seconds. This will let you to calculate the launching speed.

<u>2. Time when the speedometer displays a reading of 20 m/s</u>

First, calculate the launching speed:

  • Vf = Vo - gt

Since the ball was 3 seconds going upward and the speed at the maximum altitude is 0 you get:

  • 0 = Vo - gt

   

  • Vo = gt = 10 m/s² × 3 s = 30 m/s

Now, use the initial velocity to calculate when the ball is going upward with the speedometer reading is 20 m/s

  • 20 m/s = 30 m/s - 10 m/s² × t

  • t = [ 30 m/s - 20 m/s] / [10 m/s²] = 1 s

Thus, the first answer is t = 1 s.

<u />

<u>3. Time when the speedometer displays a reading of 30 m/s</u>

This is the same speec estimated for the launching: 30 m/s.

So, this reading corresponds to the moment when the ball was launched.

Thus time is 0, i.e. it is the same instant of the launch.

If you had worked with g = 9.80 m/s², the time had been negative. This is due to the precision of the instruments.

That is why I recommended to work with g = 10 m/s².

6 0
3 years ago
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