What Group is this element in?

Addition reactions of alkenes are characterized by _________. A) formation of a bond B) addition of two groups across a double b

mr_godi [17]

Answer:

B. ADDITION OF TWO GROUPS ACROSS A DOUBLE BOND

Explanation:

Addition reaction of alkenes involves the conversion of the double bond in alkenes Inyo single bonds by the addition of two groups of atoms or radicals.

During this addition reaction, two substances, an unsaturated compound(e.g. ethane) and an attacking reagent (hydrogen, halogens, hydrogen halides, chlorine and bromine water) combines to form a single new compound without forming any other products. So a saturated product or one in which is an increase in degree of saturation is formed.

7 0

2 years ago

For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i

frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
The charge on simple ions signifies their oxidation number.
The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

K + (-1) + 2(-2) = 0

K-5 = 0

K = +5

The oxidation number of K in KCl:

K + (-1) = 0

K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0

2 years ago

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For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

2 years ago

If an automobile air bag has a volume of 11.5 l , what mass of nan3 (in

True [87]

Moles of NaN3 at STP = volume of gas / 22.4 = 11.5/22.4 = 0.5mole. Massof NaN3 = moles of NaN3 x molecular weight = 0.5 x 65 = 32.5 grams.

8 0

2 years ago

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What's the liquid that passes through a filter?

kirza4 [7]

Filtration can be used to separate an insoluble solid from a liquid, or a precipitate from the reaction mixture in which it formed. The solid which collects in the filter paper<span> is called the residue. The clear liquid which passes through the </span>filter paper<span> is called the filtrate.</span>

6 0

2 years ago