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Wewaii [24]
3 years ago
12

A 65 kg diver jumps off a 10 m platform.

Chemistry
1 answer:
kramer3 years ago
8 0

Answer:

1. 6.37 KJ

2. 3.185 KJ

Explanation:

Gravitational potential energy is a form of potential energy with respect to the force of gravity.

Gravitational potential energy = mass x acceleration due to gravity x height

i.e   PE = m x g x h

1. At the top of the platform; mass = 65 kg, acceleration due to gravity = 10 m/s^{2}, and height = 10 m.

Gravitational potential energy = 65 x 9.8 x 10

                                                  = 6370 Joules

The gravitational potential of the diver at the top of the platform is 6.37 KJ.

2. Halfway down; mass = 65 kg, acceleration due to gravity = 10 m/s^{2}, and height = 5 m.

Gravitational potential energy = 65 x 9.8 x 5

                                                  = 3185 Joules

The gravitational potential energy of the diver halfway down is 3.185 KJ.

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273 K

Explanation:

The temperature of a gas at STP is 273 K. This is equal to 0°C or 32°F.

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Which describes Ernest Rutherford’s experiment?
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3. Suppose you wanted to design an experiment to test the composition of a mixture that includes sodium phenoxide (NaC6H5O). You
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Answer:

21.5mL of a 0.100M HCl are required

Explanation:

The sodium phenoxide reacts with HCl to produce phenol and NaCl in a 1:1 reaction.

To solve this question we need to find the moles of sodium phenoxide. These moles = Moles of HCl required to reach equivalence point and, with the concentration, we can find the needed volume as follows:

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<em>Moles NaC6H5O -116.09g/mol-</em>

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3 years ago
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sampl
Rzqust [24]

Answer:

0.193 L

Explanation:

Step 1:

Data obtained from the question.

Initial Volume (V1) = 0.250 L

Initial temperature (T1) = 27°C

Initial pressure (P1) = 0.850 atm

Final volume (V2) =?

Final temperature (T2) = 0°C

Final pressure (P2) = 1.00 atm

Step 2:

Conversion of celsius temperature to Kelvin temperature.

Temperature (Kelvin) = temperature (celsius) + 273

Initial temperature (T1) = 27°C = 27°C + 273 = 300K

Final temperature (T2) = 0°C = 0°C + 273 = 273K

Step 3:

Determination of the new volume of the sample of ammonia gas.

The new volume can be obtain by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

0.850x0.250/300 = 1xV2/273

Cross multiply to express in linear form

300 x V2 = 0.850x0.250x273

Divide both side by 300

V2 = (0.850x0.250x273) /300

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