Which type of bond is present within a water molecule

An atom has atomic number of 34. how many protons and electrons does it have? what is its symbol? 34 protons and electrons. se l

AURORKA [14]

Se: Selenium

Protons: 34
Electrons: All atoms in the periodic table are neutral until changed otherwise
Neutrons: 44
Atomic Mass: 78.09

>Symbol attached<

4 0

3 years ago

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ?C. The initial concentrations of Pb2+ and Cu2+ are

VARVARA [1.3K]

Answer:

a) Ecell = 0.5123 V

b) Ecell = 0.4695 V

c) [Pb2 +] = 4.75 M

Explanation:

a)

The reaction at the cathode is represented as follows:

Cu2 + + 2e- -> Cu (s) Eocathode = 0.34 V

The reaction at the anode is equal to:

Pb (s) -> Pb2 + + 2e- Eoanode = -0.13 V

The number of moles of the electrons that are involved is equal to n = 2

Standard cell potential equals Eo = Eocathode - Eoanode = 0.34 V- (-0.13 V) = 0.47 V

The initial cell potential can be calculated with the following formula:

Ecell = Eocell - - 0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (0.052 / 1.4) = 0.5123 V

b)

The reaction in the cell is equal to:

Cu2 + + Pb (s) -> Cu (s) + Pb2 +

The concentration of Cu2 that gives the exercise is equal 0.2 M

Therefore, the change in concentration for Cu2 + is equal to:

Cu2 + = 1.4 M - 0.2 M = 1.2 M

We use the formula from part a)

Ecell = Eocell - (0.0592 / n) log ([(Pb2 +)] / [(Cu2 +)]) = 0.47 - (0.0592 / 2) log (1,252 / 1.2) = 0.4695 V

c)

To find the concentration of Pb2 + when there is a potential change in the cell of 0.37 V, we must clear the concentration of Pb2 + from the following formula:

Eccell = Echocell - (0.0592 / n) log (([Pb2 +]) / ([Cu2 +]))

0.0296 log ([Pb2 +] / [Cu2 +]) = (Eocélula - Ecélula / 0.0296)

Clearing Pb2 +:

[Pb2 +] = 4.75 M

8 0

3 years ago

1. Compare and contrast the characteristics of metals and nonmetals.

Alona [7]

Answer:

1.Metals

These are very hard except sodium

These are malleable and ductile pieces

These are shiny

Electropositive in nature

Non-metals

These are soft except diamond

These are brittle and can break down into pieces

These are non-lustrous except iodine

Electronegative in nature

2. The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents.In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative.

3. Arrhenius theory, theory, introduced in 1887 by the Swedish scientist Svante Arrhenius, that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H+), and that bases ionize in water to yield hydroxide ions (OH−).

4. The common application of indicators is the detection of end points of titrations. The colour of an indicator alters when the acidity or the oxidizing strength of the solution, or the concentration of a certain chemical species, reaches a critical range of values.

4 0

3 years ago

Read 2 more answers

The ratio of the specific heat of aluminum to the specific heat of iron is 2:1. How much energy must be transferred to the alumi

erastova [34]

Given what we know about ratios, we can confirm that in comparison to the iron pan, the aluminum pan will receive twice as much heat.

<h3>Why would the aluminum pan receive double the heat?</h3>

This has to do with the ratio being presented.
A ratio of 2:1 means that for every unit of heat to the iron, the aluminum receives 2.
In other words, the aluminum pan receives twice as much heat when compared to the iron pan.

Therefore, we can confirm that because the ratio of heat to the aluminum in comparison to the iron pan is 2:1, this means that for every 1 unit of heat to the iron pan, the aluminum pan will receive 2. This results in double the total heat received by the aluminum pan.

To learn more about ratios visit:

brainly.com/question/1504221?referrer=searchResults

3 0

3 years ago

A voltaic cell made of a Cr electrode in a solution of 1.0 M in Cr3+ and a gold electrode in a solution that is 1.0 M in Au3+.

Vikki [24]

Answer: a) Anode: $Cr\rightarrow Cr^{3+}+3e^-$

Cathode: $Au{3+}+3e^-\rightarrow Au$

b) Anode : Cr

Cathode : Au

c) $Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E_{cell}=2.14V$

Explanation: -

a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

At cathode: $Au{3+}+3e^-\rightarrow Au$

At anode: $Cr\rightarrow Cr^{3+}+3e^-$

b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.

At anode which is a negative terminal, oxidation occurs which is loss of electrons.

Gold acts as cathode ad Chromium acts as anode.

c) Overall balanced equation:

At cathode: $Au{3+}+3e^-\rightarrow Au$ (1)

At anode: $Cr\rightarrow Cr^{3+}+3e^-$ (2)

Adding (1) and (2)

$Au^{3+}+Cr\rightarrow Au+Cr^{3+}$

d) $E^0_(Cr^{3+}/Cr)$ = -0.74 V