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slega [8]
2 years ago
15

What branch of chemistry studies the flow of electrons?

Chemistry
1 answer:
grin007 [14]2 years ago
6 0

Answer:

Your answer is B, Electrochemistry!

Explanation:

This is the part of chemistry that studies the chemical process in which electrons flow. This flow is called electricity. Electricity is generated by the flow of electrons, from one element to another element. This reaction is called oxidation reduction.

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What is the equilibrium expression for the reatcion bellow<br><br> 2SO3(g)&lt;=&gt;O2(g)+2SO2(g)
Luda [366]

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

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3. A girl<br> ran 105 metres in 15 seconds. What was her speed
sweet [91]

Answer:

7 meters per second was her speed

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2 years ago
A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas.
SOVA2 [1]

Answer:

30.05 mol

Explanation:

solving the proportion

V1 / n1 = V2 / n2

5.120 L            18.10 L

–––––––– = ––––––

8.500 mol         x

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6 0
3 years ago
A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9
Ronch [10]

Answer:

The free energy = -20.46 KJ

Explanation:

given Data:

Pb²⁺ = 0.750 M

Br⁻ = 0.232 M

R = 8.314 Jk⁻¹mol⁻¹

T = 298K

The Gibb's free energy is calculated using the formula;

ΔG = ΔG° + RTlnQ -------------------------1

Where;

ΔG° = standard Gibb's freeenergy

R = Gas constant

Q = reaction quotient

T = temperature

The chemical reaction is given as;

Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)

The ΔG°f are given as:

ΔG°f (PbBr₂)  = -260.75 kj.mol⁻¹

ΔG°f (Pb²⁺)   = -24.4 kj.mol⁻¹

ΔG°f (2Br⁻)    = -103.97 kj.mol⁻¹

Calculating the standard gibb's free energy using the formula;

ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)

Substituting, we have;

ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]

ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)

      = -260.75 + 232.34

     = -28.41 kj

Calculating the reaction quotient Q using the formula;

Q = 1/[Pb²⁺ *(Br⁻)²]

   = 1/(0.750 * 0.232²)

  = 24.77

Substituting all the calculated values into equation 1, we have

ΔG = ΔG° + RTlnQ

ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)

     = -28.41 +7.95

    = -20. 46 kJ

Therefore, the free energy of reaction = -20.46 kJ

8 0
3 years ago
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Answer:

Large change in temperature makes heat flow fast.

Explanation:

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