Given R=0.0821L⋅atmmol⋅K, how many moles of gas are in a 1.25 L container at 300K and 2.07 atm?(1 point)

Identify the location of each element group on the periodic table.

Gwar [14]

What? sorry i’m confused

8 0

3 years ago

Predict the precipitate produced by mixing an Al(NO3)3 solution with a NaOH solution. Write the net ionic equation for the react

weqwewe [10]

Al(NO3)3(aq) + 3NaOH(s) --> Al(OH)3 (s) + 3NaNO3 (aq)

The precipitate here is Al(OH)3 (s), since the solid reactant is the precipitate in the aqueous solution. Usually, it is okay to assume in basic chemistry that the transition metal is going to be part of the compound that is the precipitate, especially in an acidic salt and a strong base reaction that we have here.

4 0

3 years ago

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

$E^o_{[Cu^{2+}/Cu]}=0.34V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

5 0

2 years ago

7. The nuclear model and the solar system model both show atoms with electrons circling around

Keith_Richards [23]

These models are similar because "both the nuclear model and the solar system model have concept of nucleus, electron, positively charged nucleus at center and electrons orbiting around the nucleus".

<u>Explanation:</u>

The points which showcase similarities among both the atomic structure and the composition of the solar system are like the large percentage of the mass centered in the core which is the nucleus or the sun. Attractive forces bring the structure together by application of electromagnetic force or gravitational force.

The existence of the external objects influences the stable state of other surrounding objects as suggested by Pauli theory of exclusion and gravitational disturbance. Overall on the basis of the view only, in both the model their is similarity like atomic model speaks about electrons orbiting the nucleus similar to planets orbiting the sun.

6 0

3 years ago

Calculate the mass of one atom of helium.

mixer [17]

Answer:

One atom lf helium has a mass of 4 u

Know that 1 u =1.66. 10-24g

8 0

3 years ago