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Schach [20]
3 years ago
12

Helpppppppppppppppppppppppp

Chemistry
1 answer:
Greeley [361]3 years ago
3 0
B, the height above or below sea level
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Due to radioactive decay over many millions of years, it is possible for the element uranium to decay in
ruslelena [56]

The uranium within these items is radioactive and should be treated with care. Uranium's most stable isotope, uranium-238, has a half-life of about 4,468,000,000 years. It decays into thorium-234 through alpha decay or decays through spontaneous fission.

3 0
3 years ago
Which of the following is an example of an abiotic factor? A wind factor B Human C bacterium D duck
solmaris [256]
Abiotic factors, also called abiotic components, are non-living parts of the environment which affects the living organisms and their function in the ecosystem. I think the correct answer is A. Wind factor is the only non-living part from the choices.

7 0
3 years ago
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10. An atom that has 13 protons and 13 electrons is a neutral aluminum (Al) atom. An atom that has 13 protons and 10 electrons i
kotegsom [21]

Answer:

An atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.

For example,

Al atom has 13 protons and 13 electrons. The number of positive and negative charge is equal thus it will be neutral atom.

While the atom of Al which have 13 proton and 10 electron is not neutral. The positive charge is greater than negative by 3. Which means 3 electrons are lose by Al atom and form cation "Al⁺³".

Thus an atom of Al which has 13 protons and 10 electrons is Al cation (Al⁺³)

3 0
3 years ago
Please helpppppppppppppppp
blagie [28]

Answer:

False

Explanation:

3 0
3 years ago
Read 2 more answers
A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final
VMariaS [17]
M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C

m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C

T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
7 0
3 years ago
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