What is the shape of pure carbon fullerenes?

Given the molecules diethyl ether (CH₃CH₂OCH₂CH₃) and 1-butanol (CH₃CH₂CH₂CH₂OH), __________ has the higher boiling point mainly

MArishka [77]

Answer:

1-butanol has higher boiling point mainly due to presence of hydrogen bonding.

Explanation:

Diethyl ether is a polar aprotic molecule due to presence of polar C-O-C moiety. Hence only dipole-dipole intermolecular force exist between diethyl ether molecules.

1-butanol is a polar protic molecule due to presence of C-OH moiety. Therefore dipole-dipole force along with hydrogen bonding exist between 1-butanol molecules.

So, intermolecular force is higher in 1-butanol as compared to diethyl ether. Hence more temperature is required to break intermolecular forces of 1-butanol to boil as compared to diethyl ether.

So, 1-butanol has higher boiling point mainly due to presence of hydrogen bonding.

3 0

2 years ago

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A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the

Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

Initial temperature of the sample = 95.0 °C

Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q = m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C

-4784 * c(aluminium) = -4184

c(aluminium) = 0.875 J /g°C

The specific heat of aluminium is 0.875 J/g°C

7 0

2 years ago

How many ml of 12.0 M H2S04 are needed to make 500.0 ml of a 1.00 M solution? What happens to the freezing point and the boiling

lora16 [44]

Answer:

41.66 mL of 12.0 M sulfuric acid are needed.

Explanation:

Concentration of sulfuric acid solution taken = $M_1=12.0M$

Volume of the 12.0 M Solution = $V_1$

Concentration of required solution = $M_2=1.00M$

Volume of required 1.00 M solution = $V_2=500.0 mL$

$M_1V_1=M_2V_2$ (Dilution)

$V_1=\frac{1.00 M\times 500.0 mL}{12.0 M}=41.66 mL$

41.66 mL of 12.0 M sulfuric acid are needed.

The freezing point and the boiling point of a solvent when a non-volatile solute is dissolved in it decrease and increase respectively.

3 0

3 years ago

I need 3 facts about each and similarities plz

k0ka [10]

Mitosis- asexual reproduction
Quick
for reproductive purposes

Meiosis-sexual reproduction
Takes longer
For repairing

3 0

2 years ago

What is the role of neutrons in the nucleus of an atom?

zhenek [66]

Neutrons keep the Protons "in check", meaning Protons hold a very strong repulsive positive charge. The Neutrons counteract the repulsive force within a small space to keep the Nucleus stable.

I hope this helps! :)

4 0

3 years ago