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3 years ago

[O.01]Which of these is the location where sea floor spreading occurs? abyss guyot ocean trenches mid-ocean ridges

Chemistry

1 answer:

Alika [10]3 years ago

6 0

Answer:

Sea floor spreading occurs in mid-ocean ridges

Explanation:

mid-ocean ridge: A location where sea floor spreading occurs. The release of magma at these active volcanic sites forms new rocks and rows of mountains.

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Which of these agreements has goals of peace and shared scientific research?

Jet001 [13]

Option C, Antarctic Treaty System, is the right answer.

The Antarctic Treaty System, control global connections with regard to Antarctica; the only continent of the earth without the native population of human beings. In other words, the ATS is the entire system of adjustments developed for the intention of managing associations among states in the Antarctic. The main goal of the ATS is to guarantee "in the affair of all humankind that Antarctica shall proceed always to be practised completely for peaceful objectives and shall not become the view or gadget of universal disharmony.

8 0

3 years ago

Read 2 more answers

Please help me #6!!!!!!!!

Vesnalui [34]

I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12

3 0

3 years ago

A 1.25 g gas sample occupies 663 ml at 25∘ c and 1.00 atm. what is the molar mass of the gas?

lakkis [162]

Using ideal gas equation,

$P\times V=n\times R\times T$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values at STP will be:

P=1 atm

T=25 C+273 K =298.15K

V=663 ml=0.663L

R=0.0821 atm L mol ⁻¹

Mass of gas given=1.25 g g

Molar mass of gas given=?

$Number of moles of gas, n= \frac{Given mass of the gas}{Molar mass of the gas}$

$Number of moles of gas, n= \frac{1.25}{Molar mass of the gas}$

Putting all the values in the above equation,

$1\times 0.663=\frac{1.25}{Molar mass of the gas}\times 0.0821\times 298.15$

Molar mass of the gas=46.15

3 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

3 years ago

g An analytical chemist is titrating of a solution of diethylamine with a solution of . The of diethylamine is . Calculate the p

Serga [27]

Answer:

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Explanation:

5 0

3 years ago