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KatRina [158]
4 years ago
11

An aluminum can of a soft drink is placed in a freezer. later, you find that the can is split open and its contents frozen. work

was done on the can in splitting it open.
Chemistry
1 answer:
katen-ka-za [31]4 years ago
5 0
This is a true statement, if thats your inquiry.
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If the theoretical yield of the reaction below corresponds to 25.3 g and the percent yield of the reaction is known to be reprod
maksim [4K]

Answer:

Actual yield = 20.52 g

Explanation:

Given data:

Theoretical yield = 25.3 g

Percentage yield = 81.1 %

Actual yield = ?

Solution:

Formula:

Percentage yield = Actual yield /theoretical yield × 100

81.1 % /100 = Actual yield / 25.3 g

0.811 = Actual yield / 25.3 g

Actual yield = 0.811 × 25.3 g

Actual yield = 20.52 g

3 0
3 years ago
Part A
katovenus [111]

Answer:1.7

Explanation:

6 0
3 years ago
What evidence do you have that the slope of the natural logarithm plot will be the same for all ions of the same charge, and not
dimulka [17.4K]

Answer:

See explaination

Explanation:

1) As Evidence we just need to observe the nernst equation,

E = E0 + 2.303 RT / nF log QR

E = cell potential

E0 = Standard cell potential

R = Universal gas constant 8.314 J/K Mol

n = Valency of ion

F = faradays constant 96485 C/ mol

T = temperature in Kelvin

QR = Reaction quocient

If we plot a graph between E and Concentration of ion , We get a straight line according to the equation y = mx + C. that is the nernst equation is in the form of equation of a straight line.

E = E0 + 2.303 RT / nF log QR

Here the slope will be 2.303 RT / nF.

That is if we are measuring the electrode potential at a constant temperature , the only factor which affects the slope will be the the charge of ions 'n'. If n is same for all the ions used the slope of natural logaritham plot wil be same.

if n = 1, at room temperature ( 273 + 25 = 298 K)

2.303 * 8.134 * 298 / 1 * 96486 = 0.0591

Thus it can be concluded that the slope of natural logarithm plot will be same for all ions of the same charge.

2) The Zn metal undergo Oxidation,

Zn(s) \rightarrow Zn2+ (Aq) + 2 e-

Copper undergo Reduction

Cu2+ (Aq)+ 2 e-\rightarrow Cu (s)

The standard electrode potential for the reaction is 1.10 V

Here what happens to Zinc is dissolution into the solution , that is oxidation and what happens at copper is deposition. As Zinc ions goes into the solution , accumulation of Zinc ions around the zinc electrode takes place. ant the electron realeased into the solution reaches to the copper solution through the wire connected between them. copper ions accepts the electron and turns into copper atom (Reduction) and gets deposited at the copper electrode. This electron flow is the current generated.

As copper ions gets reduced and deposits on the copper electrode the copper ions in the solution decreases. Ultimately a situation comes where the zn ions dissolution stops because the zn ions surrownding the anode repells the zn electrode and oxidation stops. At the cathode the absence of cu2+ at cathode electrode interface stops the deposition of at copper electrode also. In short the oxidation and reduction reactions stops . This means that the cell battery power is all used up. A battrery consist of galvanic cell. batteries stores energy in chemical form and deliver it as electrical energy through electrochemical reaction as explained above. when half reactions stops in the battery no current flow takes place and battery stops.

6 0
4 years ago
Molecules have Question 8 options: A) only kinetic energy. B) neither kinetic nor potential energy. C) only potential energy. D)
emmasim [6.3K]
The answer is C. only potential energy
6 0
3 years ago
Solutions of mercury (I) nitrate and potassium bromide are mixed
Afina-wow [57]
With that informatio you can:

1) Write the chemical equation
2) Balance the chemical equation
3) State the molar ratios
4) Predict if precipitation occurs.

I will do all four, for you:

1) Chemical equation:

mercury(I) nitrate  potassium bromide    mercury(I) bromide  potassium nitrate
<span>Hg2(NO3)2             +        KBr                   → Hg2Br2          +      KNO<span>3

2) Balanced chemical equation
</span></span>
<span>Hg2(NO3)2 + 2KBr → Hg2Br2 + 2KNO<span>3

3) Molar ratios or proportions:

1 mol </span></span><span>Hg2(NO3)2 : 2 mol KBr : 1 mol Hg2Br2 : 2 mol KNO<span>3

4) Prediction of precipitation.

You can use the solubility rules or a table of solubilities. I found in a table of solutiblities that mercury(I) bromide is insoluble and potassium bromide is soluble, Then you can predict that the precipitation of mercury(I) bromide will occur.


</span></span>
7 0
3 years ago
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