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insens350 [35]
3 years ago
6

Why does the large number of hydrogen atoms in the universe suggest that other elements?

Physics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

Explanation:

The abundance of the chemical elements is a measure of the occurrence of the chemical elements relative to all other elements in a given environment. Abundance is measured in one of three ways: by the mass-fraction (the same as weight fraction); by the mole-fraction (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the volume-fraction. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and ideal gas mixtures. Most abundance values in this article are given as mass-fractions.

For example, the abundance of oxygen in pure water can be measured in two ways: the mass fraction is about 89%, because that is the fraction of water's mass which is oxygen. However, the mole-fraction is about 33% because only 1 atom of 3 in water, H2O, is oxygen. As another example, looking at the mass-fraction abundance of hydrogen and helium in both the Universe as a whole and in the atmospheres of gas-giant planets such as Jupiter, it is 74% for hydrogen and 23–25% for helium; while the (atomic) mole-fraction for hydrogen is 92%, and for helium is 8%, in these environments. Changing the given environment to Jupiter's outer atmosphere, where hydrogen is diatomic while helium is not, changes the molecular mole-fraction (fraction of total gas molecules), as well as the fraction of atmosphere by volume, of hydrogen to about 86%, and of helium to 13%.[Note 1]

The abundance of chemical elements in the universe is dominated by the large amounts of hydrogen and helium which were produced in the Big Bang. Remaining elements, making up only about 2% of the universe, were largely produced by supernovae and certain red giant stars. Lithium, beryllium and boron are rare because although they are produced by nuclear fusion, they are then destroyed by other reactions in the stars.[1][2] The elements from carbon to iron are relatively more abundant in the universe because of the ease of making them in supernova nucleosynthesis. Elements of higher atomic number than iron (element 26) become progressively rarer in the universe, because they increasingly absorb stellar energy in their production. Also, elements with even atomic numbers are generally more common than their neighbors in the periodic table, due to favorable energetics of formation.

The abundance of elements in the Sun and outer planets is similar to that in the universe. Due to solar heating, the elements of Earth and the inner rocky planets of the Solar System have undergone an additional depletion of volatile hydrogen, helium, neon, nitrogen, and carbon (which volatilizes as methane). The crust, mantle, and core of the Earth show evidence of chemical segregation plus some sequestration by density. Lighter silicates of aluminum are found in the crust, with more magnesium silicate in the mantle, while metallic iron and nickel compose the core. The abundance of elements in specialized environments, such as atmospheres, or oceans, or the human body, are primarily a product of chemical interactions with the medium in which they reside.

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A baseball on a T-ball stand has no momentum until it is hit with a bat. When Tyler swings the bat, it has a momentum of 12 kg m
NeTakaya

Answer:

4 kg m/s (option B)

Explanation:

We use conservation of momentum to solve this problem.

In the initial state, the ball has momentum zero (no momentum), and the bat has a momentum of 12 kg m/s.

Since the momentum of the system that interacts ("bat plus ball"), has to be conserved (that means the final total momentum must equal the initial total momentum), we have:

Total initial momentum = 0 kg m/s + 12 kg m/s = 12 kg m/s

Total final momentum = 8 kg m/s + X

where X is our unknown: the final momentum of the bat, and 8 kg m/s is the final momentum of the ball.

We make these two quantities equal since the total momentum has to e conserved, and solve for the unknown X in the equation:

Total initial momentum = Total final momentum

12 kg m/s = 8 kg m/s + X

12 kg m/s - 8 kg m/s = X

X = 4 kg m/s

4 0
3 years ago
The image shows the path of a ball from the time it’s thrown to the time it lands on the ground. Determine the kind of energy th
LenaWriter [7]

At Position 1: The ball has both PE and KE. (The ball is at a height from ground and velocity is given to the ball)

At Position 2: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.

At Position 3: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 4: The ball has only PE. (The ball is at a height from ground and velocity is zero.)

At position 5: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 6: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 7: The ball has only KE (Just before hitting the ground). After landing the ball have neither of the energy.

4 0
3 years ago
An atom has seven protons, eight neutrons, and six electrons. What type of electrical charge does it possess?
Crank

Answer:

positive

Explanation:

No. of positive charge= 7

No.of negative charge = 6

so 7 - 6 = 1 positive charge

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3 years ago
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vichka [17]
Solar energy
thernal energy
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3 years ago
Copper wire of diameter 0.289 cm is used to connect a set of appliances at 120 V, which draw 1850 W of power total. The resistiv
vodka [1.7K]

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W

5 0
3 years ago
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