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Fed [463]
3 years ago
9

17 POINTS (SERIOUS ANSWER ASAPP)

Chemistry
1 answer:
katovenus [111]3 years ago
6 0
I think the answer is C. Hope this helped. 
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A concentration cell is constructed by using the same half-reaction for both the cathode and anode. What is the value of standar
Akimi4 [234]

Solution :

A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.

We know,

In a standard cell,

the reduction half cell reaction is :

$Ag^+(aq)+e^- \rightarrow Ag(s) E^0 = -0.80 \ V$

The oxidation half ell reaction :

$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$

Thus the complete reaction of the cell is :

$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$

$E^0 $ cell = $E_R - E_L = 0.00  \ \text{volts}$

7 0
3 years ago
True or false, when a material loses thermal energy, it usually contracts.
Brums [2.3K]

Answer:

True

Explanation:

Whenever something loses thermal energy, it contracts and becomes smaller.

8 0
3 years ago
Read 2 more answers
At what temp will a gas be at if you allow it to expand from an original 456 mL to 65°C to 3.4 L
alexira [117]
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
                                          V₁T₂ = V₂T₁
Substituting the known values,
                                (0.456 L)(65 + 273.15) = (3.4 L)(T₁)
                                             T₁ = 45.33 K
5 0
3 years ago
Read 2 more answers
Consider the equilibrium reaction and its equilibrium constant expression. Br 2 ( g ) + 2 NO ( g ) − ⇀ ↽ − 2 NOBr ( g ) K = [ NO
zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

2 Br_2 ( g ) + 4 NO ( g ) \rightleftharpoons  4NOBr ( g )

The suitable equilibrium constant turns out:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Or in terms of the initial equilibrium constant:

K_2=K_1^2

Since the second reaction is a doubled version of the first one.

Best regards.

5 0
3 years ago
What would you multiply "moles of oxygen" by to get the units "grams of oxygen"?
photoshop1234 [79]

Answer:

AM

Explanation:

to go from moles to grams you multiply by the Atomic Mass or Molar Mass (Atomic Mass for an element and Molar Mass for a compound).

The formula is:

Mass = moles * MM

6 0
3 years ago
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