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Vera_Pavlovna [14]
3 years ago
15

Which order shows the size of the inner planets from smallest to largest

Chemistry
2 answers:
elixir [45]3 years ago
4 0

Answer:

PLUTO

MERCURY

MARS

VENUS

EARTH

NEPTUNE

URANUS

SATURN

JUPITER

Explanation:

aivan3 [116]3 years ago
4 0

Answer:

  1. PLUTO
  2. MERCURY
  3. MARS
  4. VENUS
  5. EARTH
  6. NEPTUNE
  7. URANUS
  8. SATURN
  9. JUPITER

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Help can you match the answers to the definition PLZ Need this ASAP 30 points
Alchen [17]

Answer:

1-C

2-D

3-B

4-A

Explanation:

I think it is this if it is not , sorry!

8 0
3 years ago
Is the asthenosphere or the lithosphere closer to the outer core
Murrr4er [49]
The asthenosphere is directly below the lithosphere so the answer to your question is the asthenosphere, because the outer core is towards the center of Earth.
4 0
4 years ago
How many atoms are in 2.10 g of K? molar mass of K=39.10g/mole.
Alborosie

Answer:

<h3>The answer is option B</h3>

Explanation:

To calculate the number of atoms we must first calculate the number of moles

Molar mass = mass / number of moles

number of moles = mass / Molar mass

Molar mass (K) = 39.10mole

mass = 2.10g

number of moles = 2.10/ 39.10

= 0.0537mol

After that we use the formula

N = n × L

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10^23 entities

Number of K atoms is

N = 0.0537 × 6.02 × 10^13

<h3>N = 3.23×10^22 atoms of K</h3>

Hope this helps you.

5 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
4 years ago
A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water
Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be (C_6H_{12}O)_n

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0
3 years ago
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