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Molodets [167]
3 years ago
9

Select the keyword or phrase that will best complete each sentence. must undergo addition because they have easily broken tt bon

ds. rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom. are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon. have good leaving groups and therefore readily undergo substitution and elimination reactions. In hydroboration, the boron atom bonds to the substituted carbon. undergo substitution and elimination, but can do so only when the heteroatom is made into a good leaving group.
Chemistry
1 answer:
STatiana [176]3 years ago
8 0

Alkenes must undergo addition because they have easily broken tt bonds.

Markonikov's rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom.

alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.

Alkyl halides have good leaving groups and therefore readily undergo substitution and elimination reactions.

In hydroboration, the boron atom bonds to the substituted carbon.

Hydroxides, amines and alcoxides undergo substitution and elimination, but can do so only when the heteroatom is made into a good leaving group.

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The volume of a gas is increased from 445mL to 515ml. If the initial pressure was 120kPa, what is the final pressure of the gas?
Hoochie [10]

Answer: 103.7 kPa

Explanation:

Given that:

initial volume of gas V1 = 445ml

initial pressure of gas P1 = 120 kPa.

new Volume V2 = 515ml

new pressure P2 = ?

Since, only pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

120 kPa x 445ml = P2 x 515 kPa

53400 = P2 x 515 kPa

P2 = 53400/515

P2 = 103.7 kPa

Thus, the final pressure of the gas is 103.7 kPa

6 0
3 years ago
A current is applied to two electrolytic cells in series. In the first, silver is deposited; in the second, a zinc electrode is
vitfil [10]

The amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Given ,

A current is applied to two electrolytic cells in series .

In the first ,silver is deposited and in the second a zinc electrode is consumed .

the reactions involving are ;

Ag+ (aq) + e = Ag

Zn = Zn2+ (aq) +2e

thus the resultant equation is ,

2Ag+ (aq) +Zn = 2Ag + Zn2+

Thus for every mole of Zn dissolves , there is 2 moles of Ag is formed .

65.38 g of Zn contains = 1 moles

1.2 g of Zn contains = 1.2/65.38 =0.01835 moles

for every 1 mole of Zn dissolves there is 2 moles of Ag formed .

Thus the amount of Ag formed in moles =2(O.01835) =0.0367 Moles

1 mole of Ag contains = 107.86g

0.0367 moles of Ag contains = 107.86 (0.0367) =3.959 g of Ag

Hence ,the amount of Ag plated out if 1.2 g of Zn dissolves is 3.959 g .

Learn more about electrolytic cell here:

brainly.com/question/19854746

#SPJ4

6 0
1 year ago
How many moles of naf are in 34.2 grams of a 45.5% by mass naf solution?
Verdich [7]
 <span>% by mass = mass solute x 100 / mass solution 

45.5 % = mass solute NaF x 100 / 34.2 

mass solute NaF = 34.2 x 45.5 /100=15.6 g 

molae solute NaF = 15.6 g ( 1 mol / 41.9887 g)= 0.372</span>
8 0
3 years ago
What did thomson’s model of the atom called
nignag [31]

Answer:

Plum pudding model

Explanation:

Thomson’s model of the atom called: PLUM PUDDING MODEL

8 0
3 years ago
Read 2 more answers
A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).
morpeh [17]

Answer:

20.0

Explanation:

NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL

8 0
3 years ago
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