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Ray Of Light [21]
3 years ago
10

At what velocity (m/s) must a 19.9 g object be moving in order to possess a kinetic energy of 1.0 J

Chemistry
2 answers:
solmaris [256]3 years ago
7 0

Answer:

for this one you would have to divide 19.9g by the kinetic energy 1.0 and you would get your answer.

Explanation:if you would give me brainiest that would help a lot :)

slamgirl [31]3 years ago
4 0

Answer:

Explanation:

T

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kondor19780726 [428]

Answer: gravity affect the amount of both kinetic and potential energy

Explanation:

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4 years ago
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.The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temper
Elis [28]

Explanation:

It is given that the initial mass of benznene is 7.9286 g

Mass of benzene left = 5.9987 g

So, mass of benzene with which gas get saturated will be calculated as follows.

               = 7.9286 g  - 5.9987 g  = 1.9299 g

Therefore, moles of benzene with which gas get saturated = \frac{mass}{ molar mass}

                = \frac{1.9299 g}{78.112 g/mol}

                = 0.0247 moles

Temperature = 27.3^{o}C = 27.3 + 273.15 = 300.45 K

Volume = 5.01 L

So, according to ideal gas equation PV = nRT

Putting the given values into the ideal gas equation as follows.

                       PV = nRT

           P \times 5.01 L = 0.0247 mol \times 62.36 torr-liter/mol K \times 300.45 K

                   P = \frac{462.781 torr-liter}{5.01 L}

                      = 92.371 torr

Hence, we can conclude that vapor pressure of benzene is 92.371 torr.

8 0
3 years ago
Gordon throws a baseball into the air. It rises, stops when it reaches its greatest height, and then falls back to the ground. A
Ivahew [28]

That would be when the baseball is rising. The energy from the throw is kinetic energy.

7 0
3 years ago
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A gas has a volume of 590 mL at a temperature of –55.0°C. What volume will the gas occupy at 30.0°C? Show your work.
Mashutka [201]
Step 1: Convert the temperatures to K
-55 + 273 = 218 K = T1
30 + 273 = 303 K = T2

Step 2: Use the formula V1/T1 = V2/T2
590/218 = V2/303
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8 0
3 years ago
1.00 x 10^6 atoms of gold is equivalent to how many grams?
S_A_V [24]

Answer:

3.27 x 10⁻¹⁶ grams

Explanation:

moles Au = 1.00 x 10⁻⁶ Atoms / 6.02 x 10²³Atoms / mole = 1.66 x 10⁻¹⁸ mole Au

grams Au = 1.66 x 10⁻¹⁸ mole Au x 196.97 grams Au/mole Au

= 3.27 x 10⁻¹⁶ grams Au

8 0
4 years ago
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