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Colt1911 [192]
3 years ago
8

What is the primary source of heat in the central cores of the giant planets?

Chemistry
2 answers:
Firlakuza [10]3 years ago
6 0

Answer:

ExplEach giant planet has a core of “ice” and “rock” of about 10 Earth masses. Jupiter, Saturn, and Neptune have major internal heat sources, obtaining as much (or more) energy from their interiors as by radiation from the Sun.

What are the main atmospheric heat sources of each of the giant planets? For Jupiter, Saturn, and Neptune, both sunlight and internal sources provide energy to the atmosphere.

Yanka [14]3 years ago
3 0

Answer:

Each giant planet has a core of “ice” and “rock” of about 10 Earth masses. Jupiter, Saturn, and Neptune have major internal heat sources, obtaining as much (or more) energy from their interiors as by radiation from the Sun.

Explanation:

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What kind of group 1 atoms are the most reactive? a. Small atoms c. Atoms with the most electrons in the valence level b. Large
garri49 [273]

Answer:

large atoms

Explanation:

its kinda obvious

4 0
3 years ago
Read 2 more answers
What is the total number of oxygen atoms on the right-hand side of this chemical equation?
vivado [14]

The chemical equation is

Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)

Answer:

12

Explanation:

In the right hand side of the equation, there are three compound which contains O2, which are;

Cu(NO3)2 , number of oxygen atoms =3*2 =6

2NO2, number of oxygen atoms = 2*2=4

2H2O, number of oxygen atoms =2*1=2

Total number of oxygen atoms on the right side of equation = 6+4+2 =12

8 0
3 years ago
Please Help Me!!!
sergejj [24]

Answer:

The products are carbon dioxide and water

Explanation:

Step 1: Data given

Combustion = a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve  O2  as one reactant.

Step 2: The complete combustion of C3H7OH:

For the combustion of 1-propanol, we need O2.

The products of this combustion are CO2 and H2O.

C3H7OH + O2→ CO2 + H2O

On the left side we have 3x C (in c3H7OH), on the right side we have 1x C (in CO2). To balance the amount of C, we have to multiply CO2 on the right side by 3

C3H7OH + O2→ 3CO2 + H2O

On the left side we have 8x H (in C3H7OH) and 2x on the right side (in H2O). To balance the amount of H, we have to multiply H2O, on the right side by 4.

C3H7OH + O2→ 3CO2 + 4H2O

On the left side we have 3x O (1x in C3H7OH and 2x in O2), on the right side we have 10x O (6x in CO2 and 4x in H2O).

To balance the amount of O on both sides, we have to multiply C3H7OH by 2, multiply O2 by 9. Then we have to multiply 3CO2 by 2 and 4H2O by 2. Now the equation is balanced.

2C3H7OH + 9O2→ 6CO2 + 8H2O

For 2 moles propanol, we need 9 moles of O2 to produce 6 moles of CO2 and 8 moles Of H2O

The products are carbon dioxide and water

7 0
3 years ago
An element is a pure substance in which there are how many kinds of atoms?
Ostrovityanka [42]

Answer:

One

Explanation:

An element is a pure substance in which there are only one kind of atom. Elements are distinct substances that cannot be split up into simpler substances.

Such substances consists of only one kind of atom. There are over a hundred known elements to date.

Generally, as a pure substance, the composition of an element is definite and they are homogenous in all parts.

8 0
3 years ago
Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of
Oxana [17]
<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Al→3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>
7 0
4 years ago
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