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Brut [27]
3 years ago
8

If you have 2 moles of an element, how many particles do you have?

Chemistry
1 answer:
Amanda [17]3 years ago
3 0

Answer:

1.204×10^24 particles of element

Explanation:

1mol=6.02×10^23 number of particles (in this case an element)

2mol=(6.02×10^23)×2 particles of the element

=1.204×10^24 particles of element

*be careful aware significant digits

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Answer:

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Making food from light- Photosynthesis

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All living units give out power for a given time from food small complex units through formed of small units breathing and or fermentation. Some units make food small complex units using light power for a given time through the process of (making food from light).

6 0
2 years ago
What is the variable that is changed by
denpristay [2]

Answer:

b. independent/manipulated variable

Explanation:

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3 0
2 years ago
During region C of the graph above the energy being absorbed is a. increasing the velocity of the water particles. b. increasing
Mkey [24]

Answer:

Your answer would be B. Increasing the distance between the water molecules and increasing their phase energy.

Explanation:

Hope this helps you answer this question!

7 0
2 years ago
Blank is how strong gravity pulls on an object
zheka24 [161]
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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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