All categories

2 years ago

What is a double bond? PLEASE HELP

Chemistry

2 answers:

Mademuasel [1]2 years ago

8 0

Answer:

Explanation:

a covalent bond between two atoms involving four bonding electrons

sleet_krkn [62]2 years ago

4 0

Answer:

option c is correct

Explanation:

Hope it helps you......

You might be interested in

A testable explanation that's supported or refuted is called an

N76 [4]

It is called a hypothesis.

5 0

3 years ago

A solution contains one or more of the following ions: Ag + , Ca 2 + , and Co 2 + . Ag+, Ca2+, and Co2+. Lithium bromide is adde

Eva8 [605]

Answer:

Since with LiBr no precipitation takes place. So, Ag+ is absent

When we add Li2SO4 to it, precipitation takes place.

Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate

Thus, Ca2+ is present.

When Li3PO4 is added, again precipitation takes place.Reaction is:

Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate

A. Ca2+ and Co2+ are present in solution

B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)

C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)

8 0

2 years ago

What do we call the type of chemical reaction that occurs in this step? Explain how the name of this type of reaction relates to

icang [17]

Figure it out yourself

6 0

2 years ago

Cual es el propósito de la línea escalonada gruesa de la tabla periódica

noname [10]

Separa los metales de los no metales. Agregame como amiga, saludos.

3 0

3 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

Learn more:

5 0

2 years ago