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2 years ago

Deposits of gold can be found in the deep ocean. Please select the best answer from the choices provided T F

Physics

2 answers:

Igoryamba2 years ago

7 0

In some cases true, and in others false. but it is true

Setler [38]2 years ago

5 0

Answer: True

The excavation of gold from the deep oceanic vents made possible from the Atlantic and north Pacific oceans. Approximately 100 metric million tons of gold has been excavated from these oceans. Also, the undissolved gold in the seafloor has also been excavated. But currently, gold mining from the sea floor and the deep oceanic crust is not considered as efficient mining process.

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3. To open a soda bottle, a force of 55.0 N is applied to the bottle opener. If the bottle opener

Bogdan [553]

Answer:

0.0815

Explanation:

Mechanical advantage is output force/input force. So all you have to do is 55.0N/675N ~ 0.0815

Hope this helped!

5 0

2 years ago

Summarize the steps you might use to carry out an investigation using scientific methods

melomori [17]

1. Make an observation
2. Form a question
3. Form a hypothesis
4. Conduct an experiment
5. Analyze the data and draw your conclusion

8 0

3 years ago

the average time required for radioactive elements, whereby they decay to half their original mass is called the "average life"

VikaD [51]

I'm not sure what your question is. But, the half life is the amount of time required for half the material to decay. For U238 this is 4.5 billion years, whilst for Fr-223 (Francium) its about 22 minutes. To calculate the time for something to decay you need to use the equation:
Mass (after time t) = Mass (initial) * (0.5)^(time/half life)
Hope this helps

4 0

3 years ago

Read 2 more answers

A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 3300 kcal o

Tema [17]

Answer

given,

heat added to the gas,Q = 3300 kcal

initial volume, V₁ = 13.7 m³

final volume, V₂ = 19.7 m³

atmospheric pressure, P = 1.013 x 10⁵ Pa

a) Work done by the gas

W = P Δ V

W = 1.013 x 10⁵ x (19.7 - 13.7)

W = 6.029 x 10⁵ J

b) internal energy of the gas = ?

now,

change in internal energy

Δ U = Q - W

Q = 3300 x 10³ cal

Q = 3300 x 10³ x 4.186 J

Q = 1.38 x 10⁷ J

now,

Δ U = 1.38 x 10⁷ - 6.029 x 10⁵

Δ U = 1.32 x 10⁷ J

6 0

2 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

4 0

2 years ago