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SSSSS [86.1K]
3 years ago
8

What is a characteristic of branched chained alkanes?

Chemistry
2 answers:
Morgarella [4.7K]3 years ago
7 0

Thanks! that is correct THE ANSWER IS C

olga2289 [7]3 years ago
4 0

C) They have more than two carbons that are bonded to only one other carbon

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Which of these is a ball stick model
Liula [17]
I believe it is the third choice from the left because it is 3D and has the "sticks", unlike the space-filling model to the left of it. Hope this helped!
8 0
3 years ago
Read 2 more answers
Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of
Serhud [2]

Answer:  0.17 m CH_3COONH_4 : Highest freezing point

0.20 m CoSO_4: Second lowest freezing point

0.18 m MnSO_4: Third lowest freezing point

0.42 m ethylene glycol: Lowest freezing point

Explanation:

Depression in freezing point  is a colligative property which depend upon the amount of the solute.

\Delta T_f=i\times k_f\times m

where,

\Delta T_f = change in freezing point

i= vant hoff factor

k_f = freezing point constant

m = molality

a) 0.17 m CH_3COONH_4

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for CH_3COONH_4, thus total concentration will be 0.34 m

b) 0.18 m MnSO_4

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for MnSO_4, thus total concentration will be 0.36 m

c) 0.20 m CoSO_4

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for CoSO_4, thus total concentration will be 0.40 m

d) 0.42 m ethylene glycol

For non electrolytes undergoing no dissociation, vant hoff factor is equal to 1 . Thus i = 1 for ethylene glycol, thus concentration will be 0.42 m

The more is the concentration, the highest will be depression in freezing point and thus lowest will be freezing point.

4 0
3 years ago
A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f
Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0
3 years ago
Why Sodium and phosphorus are in the same period, but they have different atomic radii?
Marianna [84]

Answer:

Because sodium is on the left side of the PT and phosporus is on the right side. And atomis radius decreases from left to right.

Explanation:

4 0
3 years ago
Fe2O3 + H2 --> Fe + H2O
sveticcg [70]
A) 10.0g of Iron is \frac{10.0}{55.8}=179mmol of Iron.

One mole of dihydrogen is required to form one mole of iron, hence you'll need 0.179*1.00=0.179g of dihydrogen.

B) Likewise : 2.50g of Iron is \frac{2.50}{55.8}=44.8mmol of Iron, hence we'll need 44.8*(3*16+2*55.8)=7.15g of Fe2O3.
4 0
4 years ago
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