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Ganezh [65]
3 years ago
15

If the nucleus was the size of a tennis ball the atom would be the size of....?

Chemistry
2 answers:
Ivan3 years ago
6 0

It really depends on two things, what element it is, and how many electrons it has.
slavikrds [6]3 years ago
3 0
If you could find anything just bigger than a tennis ball, you're all good, because the nucleus is almost half the size of the atom, I think! U should find this helpful, if you do, great, I'm glad that I ciyld help
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Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with
skad [1K]

A.  Decomposing water requires a high activation energy.

Explanation:

In decomposing water to release hydrogen gas to make fuel cells, the process requires a very high activation energy.

                             2H₂O ⇆ 2H₂  + O₂

 This is the overall reaction. O-H must be broken to release free hydrogen to produce hydrogen gas.

The O-H bond is a very strong force of attraction that requires a high activation energy to overcome.

  • The activation energy is the energy barrier that must be overcome before a reaction takes place.
  • The sun is a renewable source of energy.
  • Water decomposition produces useful oxygen gas needed by all life for cellular respiration.

Learn more:

Source of energy brainly.com/question/2948717

#learnwithBrainly

4 0
3 years ago
One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
3 years ago
4HCI + O2 =2H20 + Cl2
Sergio039 [100]

The rate of the backward reaction increases

Explanation:

It is evident that if the reaction is left to proceed spontaneously, the forward reaction is favored because it results in a decrease in pressure in the system (The total reactants have 5 moles and the products have 3 in total).

Increasing H₂O concentration is then reaction, therefore, stymies the forward reaction and favors the reserves reaction. This is because the reverse reaction will lead to reduced pressure.

6 0
3 years ago
What element is in group 13, period 4
Ludmilka [50]

Answer:

Gallium

Explanation:

6 0
3 years ago
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Which of the following is a unit of heat? 1. Fahrenheit 2. Energy 3. Joule 4. Celsius
ELEN [110]
The unit of heat is Joule
6 0
3 years ago
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