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zvonat [6]
3 years ago
14

Apart from the formation of ethanol, You can't use alkenes direct hydration method to produce an alcohol where the OH group is a

t the end of a chain. Using propene (CH3CH=CH2), but-1-ene (CH3CH2CH=CH2) andbut-2-ene (CH3CH=CHCH3) as examples, explain why that is.
Chemistry
1 answer:
emmainna [20.7K]3 years ago
8 0
Wait a second...I am sorry I don't know the answer but I have something to tell you.
It is not necessary that in alcohols, OH should be at the end of a chain .

It can be anywhere in the chain..
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Please help due tomorrow science
horrorfan [7]

Answer:

1. start line drawn in ink

2. start line below solvent level

Explanation:

1. the line would run up the paper. should have drawn with pencil.

2. the samples would dissolve into solvent and not run up the paper.

8 0
3 years ago
Determine the number of HC=CH, monomeric units in one molecule of polyethylene with a molar
VladimirAG [237]

polyethylene contains HC=CH units.

mass of this is 26 gram/mol

number of such units =13500/26

3 0
3 years ago
How many molecules of sugar (C6H1206) are in a mole?
Tanzania [10]
By definition, one mole (one gram molecular weight) of any substance, contains Avogadro’s number of particles; atoms if you are discussing an element, or molecules if a compound. Avogadro’s number has been determined by several methods, all of the accepted values lie within a range of +-1% about the value of 6.022045 x 10^23/gm. That is a large number, in this case approximately; 602,204,500,000,000,000,000,000 molecules of glucose.

From the web :v
8 0
3 years ago
SOMEONE PLEASE HELP
Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

= 8.16 \times 10^{ - 6} \:  kg

\boxed{density =  \frac{mass}{volume} }

Density of iron bar

=  \frac{0.0642}{8.16 \times 10^{ - 6} }

= 7870 kg/m³ (3 s.f.)

7.

\boxed{mass = density \:  \times volume}

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

= 12  \times  10^{ - 6}  \: m^{3}  \\  = 1.2 \times 10^{ - 5}  \: m ^{3}

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

=  \frac{0.151} {1.25 \times 10^{ - 5} }  \\  = 12600 \: kg/ {m}^{3}

(3 s.f.)

6 0
3 years ago
I need chemistry help! How would I set up these problems?
Bond [772]

Answer: -

1) 8.33 minutes

2) 118.39 in/ s

180.43 m/min

10.83 km/ hr

Explanation: -

Speed of light = 3 x 10⁸ m/s

Distance of the earth from the sun= 93 million miles

We know 1 million = 1,000,000

Also 1 mile = 1609 m

Distance of the earth from the sun= 93 million miles

= 93,000,000 miles.

= 1.5 x 10^{11} m

Time taken = \frac{Distance}{Speed}

= \frac{1.5 x [tex] 10^{11} m}{3 x 10⁸ m/s} [/tex]

= 500 s

= 500/ 60

= 8.33 minutes

2) Distance = 1 mile = 63360 inches

Time taken = 8.92 min

= 8.92 x 60

= 535.2 s

Speed = \frac{distance}{time}

= \frac{63360 inches}{535.2 s}

= 118.39 in/ s

Distance = 1 mile = 63360 inches = 63360 x 2.54 cm = 63360 x 2.54 x 10^{-2} m

Time taken = 8.92 min

Speed = \frac{distance}{time}

= \frac{63360 x 2.54 x [tex] 10^{-2} m}{8.92 min} [/tex]

= 180.43 m/ min

1 m = 10⁻³ Km

1 min = 1/60 hour

1 m /min = 10⁻³ km/ \frac{1}{60 hour}

= 60/1000

=0.06 km/hr

180.43 m / min = 180 x 0.06 km / hr

= 10.93 km / hr

4 0
3 years ago
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