Answer:
Size is the additional feature that determines asteroids.
Explanation:
Asteroids are small, rocky objects that orbit the sun. Although asteroids orbit the sun like planets, they are much smaller than planets. There are lots of asteroids in our solar system. Most of them live in the main asteroid belt which is a region between the orbits of Mars and Jupiter.
The answer is
6.8 * 10^-15
The explanation:
1- we have to convert all measurements to the same units:
Conversions:
when 1 m = 100 cm
and 1 m = 10^12 pm
So,
proton radius: 1.0*10^-13 cm * (1m / 100 cm) = 10^-15 m
proton volume: 4/3 * pi * r^3 = 4/3 * pi * (10^-15 m)^3 = 4.2 * 10^-45 cu. meters
and
H atom radius: 52.9 pm * (1m / 10^12 pm) = 5.29 * 10^-11 m
H atom volume: 4/3 * pi * r^3 = 4/3 * pi * (5.29 * 10^-11 m)^3 = 6.2 × 10^-31 cu. meters
So,
2- Fraction of space occupied by nucleus = proton volume / H atom volume
= (4.2 * 10^-45 cu. meters) / (6.2 × 10^-31 cu. meters)
= 6.8 * 10^-15
So, the "fraction" would be 6.8 * 10^-15 out of 1.
<u><em>Answer:</em></u>
Correct option is D.
It accelerates the reaction rates of a mixture.
<u><em>Explanation:</em></u>
It is used to speed up a reaction by lowering the activation energy.Catalysis is the backbone of many industrial processes, which use chemical reactions to turn raw materials into useful products.
<u><em>Types</em></u>
There are two types of catalyst (1) Homogeneous (2) Heterogeneous
In a heterogeneous reaction, the catalyst is in a different phase from the reactants. In a homogeneous reaction, the catalyst is in the same phase as the reactants.
Answer:
Explanation:
A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
