Ask question

Ask question

All categories

Free_Kalibri [48]

3 years ago

5

In the covalent compound CH3, the Greek prefix used to represent the anion is Answer here

1 answer:

Contact [7]3 years ago

3 0

Answer:

Tri.

Explanation:

''Tri'' is the Greek prefix used to represent the CH3 because there are three hydrogen atoms to form the covalent compound so in Greek method ''Tri'' is used for three. In Greek system, there are certain names for prefixes such as ''di'' is used for two, ''tri'' is used for three, ''tetra'' is used for four etc. This method only provides information about the number of atoms that form the covalent compound.

You might be interested in

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

_____ is the process of linking smaller molecules to form long chains of higher molecular weight.

Leya [2.2K]

What is “Polymerization” is your answer

7 0

3 years ago

What is isostasy?

NemiM [27]

Explanation:

FJjuhjjjJJJii3ii3ii3iiभबजबछसझबूईसडड

7 0

2 years ago

How many moles of h2o are produced when 3.25 moles of o2 react in 2c2h6 + 7o2 ---> 4co2 + 6h2o

Alenkinab [10]

Answer:

2.78 moles of water are produced.

Explanation:

Given data:

Number of moles of H₂O produced = ?

Number of moles of oxygen react = 3.25 mol

Solution:

Chemical equation:

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Now we will compare the moles of water with oxygen.

O₂ : H₂O

7 : 6

3.25 : 6/7×3.25 = 2.78 mol

5 0

3 years ago

How potential energy changes when two photons approach each other and fuse to form deuterium nucleus

Phoenix [80]

nbjfhgefabshcnjhsdbfvdahjscadhgvAnswer:

efrghtyrtjhryExplanation:

dfhhgfhfgjfgjfgjfgjfgj

4 0

3 years ago