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jok3333 [9.3K]
3 years ago
6

True or False? Limiting cultivation increases compaction.

Chemistry
1 answer:
solmaris [256]3 years ago
3 0
I believe the answer is true
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The number of energy levels are determined by what __________________ they are in. So, elements with 6 energy levels are found i
Anarel [89]

Answer:

group, 6 or 16

Explanation:

as the group's progress going from the left of the table to the right of the the table, the valance electrons increase. although group 14 is split up by a ladder stair steppy thingy. the ones on the top take electrons, and the ones below give electrons.

6 0
3 years ago
The majority of the mass of the atom is located in the _______. A) nucleus B) core electrons C) pions and quarks D) valence elec
NeX [460]

The majority of the mass of the atom is located in the nucleus. Remember that the nucleus contains both protons and neutrons and therefore, most of the mass of the atom.

5 0
3 years ago
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One mole of carbon will have the same number of particles as one mole of hydrogen.
AVprozaik [17]
And the answer is False,it does not have the same number with particles as the one mole of hydrogen.
3 0
3 years ago
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HELP PLEASE!! ASAP What would the mass be in kg?
fgiga [73]

Answer:

i guess answer is 0.0600

Explanation:

Here,

Density=0.791g/cm^3

Volume=75.89ml

So,

Mass=Density*Volume

=0.791g/cm^3×75.89ml

=60.02g/cm^3.ml

Expressing them in kg

60.02/1000

=0.0600 kg

4 0
3 years ago
Calculate the equilibrium constant at 25 ∘C for the reaction Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
Murrr4er [49]

Answer:

1.7 × 10 ^42

Explanation:

Using Nernst equation

E°cell = RT/nF Inq

at equilibrium

Q=K

E°cell  = 0.0257 /n Ink= 0.0592/n log K

Fe2+(aq)+2e−→Fe(s)     E∘= −0.45 V

Ag+aq)+e−→Ag(s)         E∘= 0.80 V

Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)

balance the reaction

Fe → Fe²⁺ + 2e⁻  reversing for oxidation E° = 0.45 v

2 Ag⁺ +2e⁻ → 2Ag

n = 2 moles  and K = equilibrium constant

E° cell = 0.80 + 0.45 = 1.25 V

E° cell = (0.0592 / n) log K  

substitute the value into the equations and solve for K

(1.25 × 2) / 0.0592  = log K

42.23 = log K

k = 10^ 42.23

K = 1.7 × 10 ^42

8 0
3 years ago
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