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Anarel [89]
2 years ago
11

This information is taken directly from Exercise 3 in the CHEM 111/112 Laboratory Manual. Another example of excellence in the p

resentation of data comes from a study on the cholera outbreak in London in 1854 drawn by John Snow. By overlaying the residences of the casualties on a city map that showed sources of water, the pump on Broad Street appears to be clearly implicated in the transmission of the disease." While writing a report, a student uses this source for background information about data presentation le of excellence in data presentation comes from a study of the cholera outbreak in London in 1854. John Snow created a city map that showed the residences of the casualties and sources of water. According to this map, the pump on Broad Street appears to be clearly implicated in the transmission of the disease." He uses this citation for the report: "3. Leung. T. W. Schaefer. A. ed Exercise 3: The Graphical Depiction of Scientitic Data. General Chemistry of the Texas Environment, 7th edition. 2018. 10. Was this cited correctly? O no O yes

Chemistry
1 answer:
kolbaska11 [484]2 years ago
5 0

Answer:

<h2>No</h2>

the information was not cited correctly....

Explanation:

I hope the following explanation will help you a lot.

You might be interested in
Consider the standard galvanic cell based on the following half-reactions The electrodes in this cell are and . Does the cell po
Free_Kalibri [48]

Question: The question is incomplete and can't be comprehended. See the complete question below and the answer.

Consider a galvanic cell based upon the following half reactions: Ag+ + e- → Ag 0.80 V Cu2+ + 2 e- → Cu 0.34 V

How would the following changes alter the potential of the cell?

a) Adding Cu2+ ions to the copper half reaction (assuming no volume change).

b) Adding equal amounts of water to both half reactions.

c) Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

d) Adding Ag+ ions to the silver half reaction (assume no volume change)

Explanation:

Nernst equation relates the reduction potential of an electrochemical reaction (half-cell or full cell reaction) to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation.

Reaction under consideration:

Ag+ + e- → Ag 0.80 V

Cu+2 + 2 e- → Cu 0.34 V

Clearly, Ag reduction potential is high and this indicates that it gets reduced readily which leaves Cu to oxidize. Cu+2 ions are products of reaction and Ag+ ions are reactant ions.

Nernst equation : Ecell = E°cell­ – (2.303 RT / n F) log Q    

where                            

             Ecell = actual cell potential

             E°cell­ ­​ = standard cell potential

             R = the universal gas constant = 8.314472(15) J K−1 mol−1

             T = the temperature in kelvins

              n = the number of moles of electrons transferred                                    

 F = the Faraday constant, the number of coulombs per mole of electrons:

  (F = 9.64853399(24)×104 C mol−1)

 Q = [product ion]y / [reactant ion] x

Accordingly when applied to above reaction one will get the following

= E°cell­ – (2.303 x RT / 6 F) log [Cu+2] / [Ag+]

Now the given variables can be studied according to Le Chatelier's principle which states when any system at equilibrium is subjected to change in its concentration, temperature, volume, or pressure, then the system readjusts itself to (partially) counteract the effect of the applied change and a new equilibrium is established.

a)        Adding Cu2+ ions to the copper half reaction (assuming no volume change).

Addition of Cu+2 ions increases its concentration and consequently increases the Q value which results in reduction of Ecell. In other words the addition of Cu+2 ions favors the backward reaction to maintain the equilibrium of reaction and hence the forward reaction rate decreases.

b)       Adding equal amounts of water to both half reactions.

Addition of water increases the dilution of the electrochemical cell. For weak electrolytes such as Ag+/ Cu+2 with increase in dilution, the degree of dissociation increases and as a result molar conductance increases.

c)        Removing Cu2+ ions from solution by precipitating them out of the copper half reaction (assume no volume change).

Based on Le Chatelier's principle when Cu+2 ions amount is decreased by its continuous removal from  the system the forward reaction is favored. As the Cu+2 ions is removed the system attempts to generate more Cu+2 ions to counter the affect of its removal.

d)       Adding Ag+ ions to the silver half reaction (assume no volume change)

Addition of reactant ions, i.e. Ag+ ions, will favour the forward reaction, which results in more product formation.

6 0
1 year ago
A student mixed together aqueous solutions of Y and Z. A white precipitate(solid)formed. which could not be Y and Z
maks197457 [2]
<span>(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)

</span><span>(B)hydrochloric acid + sodium hydroxide 
</span><span>HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

</span><span>(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> </span>AgCl(s) +Ca(NO3)2(aq)

<span>(D)sodium chloride + silver nitrate
</span>NaCl(aq) +  AgNO3(aq) ----> AgCl(s) +NaNO32(aq)

AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.

4 0
2 years ago
A 2.25-g sample of magnesium nitrate, mg(no3)2, contains __________ mol of this compound.
MatroZZZ [7]
  Mg(No3)2  is  calculated   as  follows

moles  =  mass/molar  mass
the  molar  mass  of  Mg(NO3)2  is =  148 g/mol

moles  is  therefore= 2.25 g /  148  g/mol= 0.0152  moles

Mg(No3)2   contain  0.0152  moles  of  the   compound 
6 0
1 year ago
what can you say about the average distance from the nuclease of an electron the 2's orbital as compared with a 3s orbital
Alecsey [184]

Answer:

A 3s orbital is at a greater average distance from the nucleus than a 2s orbital

Explanation:

As the principal quantum number n increases, the distance of the orbital from the nucleus increases. Hence if we consider the 2s and 3s orbitals, it is easy to see that the 3s orbital is at a greater distance from the nucleus than the 2s orbitals.

This is clearly seen when we plot the radial distribution against the distance from the nucleus. This enables us to visualize the region in space in which an electron may be found.

7 0
2 years ago
A system delivers 225 j of heat to the surroundings while delivering 645 j if work calculate the change in the internal Chang
Charra [1.4K]

Heat given out to the surroundings by the system = 225 J

Work done by the system on the surroundings = 645 J

According to the energy conservation, the energy can neither be created nor it can be destroyed, it can transform from one form to another. Hence, the energy which is lost to the surrounding as a work done and heat came from the internal energy of the system.

Hence, the change in the internal energy = - 225 - 645 = - 870 Joules

Negative sign means that the internal energy of the system is decreased by 870 Joules

3 0
2 years ago
Read 2 more answers
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