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3 years ago

Which letter represents the activated complex? 1. A 2. B 3. F 4. G

Chemistry

1 answer:

lana66690 [7]3 years ago

4 0

Answer:

2. B.

Explanation:

The letter B represents the activated complex.

The activated complex is the intermediate that is formed between the states of reactants "F" and products "G".

Letter A represents the activation energy of the reaction, that is the difference in potential energy between the reactants "F" and activated complex "B".

Letter C represents the enthalpy change of the reaction, that is the difference in potential energy between products "G" and reactants "F".

Letter D represents the potential energy of products "G".

Letter E represents the potential energy of reactants "E".

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When bonds between atoms are broken or formed, what is the outcome?

LuckyWell [14K]

Two or more atoms may bond with each other to form a molecule. When two hydrogens and an oxygen share electrons via covalent bonds, a water molecule is formed. Chemical reactions occur when two or more atoms bond together to form molecules or when bonded atoms are broken apart.

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3 years ago

Read 2 more answers

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

Masja [62]

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

Explanation:

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:

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

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

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

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

Give the chemical symbol for the element with the ground‑state electron configuration [ Ar ] 4 s 2 3 d 1 . symbol: Determine the

Rus_ich [418]

Answer:

Sc (Scandium) has the given electronic configuration.

Explanation:

The given electronic configuration is [Ar] $4s^{2}3d^{1}$ .

The last electron enters the d-subshell and hence is a d-block element known as Scandium with chemical symbol Sc.

For 4s subshell

n=4,l=0 and m ranges from -l to +l so m=0.

For 3d subshell

n=3,l=2 and m ranges from -l to +l so m can take values -2,-1,0,+1,+2

Note:

l values for subshells:

s : 0

p : 1

d : 2

f : 3 and so on.

5 0

3 years ago

I need help with 2, 3, and 4 please help i will give the brainliest

Sphinxa [80]

Answer:

2=2.28*10^5

3=3.98*10^3

4=5.76*10^3

8 0

3 years ago

Can anybody can solve this PLEASE

strojnjashka [21]

Answer:

Explanation:

b is the most stable( noble gas ) since it has an octet valance shell and can't loses or gains any more of electrons

8 0

3 years ago