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2 years ago

What name is given to the carbon-based structures used in lubricants and nanotubes

Chemistry

2 answers:

sergejj [24]2 years ago

8 0

Answer:

Fullerene, also called buckminsterfullerene, any of a series of hollow carbon molecules that form either a closed cage (“buckyballs”) or a cylinder (carbon “nanotubes”).

Explanation:

A fullerene is an allotrope of carbon whose molecule consists of carbon atoms connected by single and double bonds to form a closed or partially closed mesh, with fused rings of five to seven atoms. The molecule may be a hollow sphere, ellipsoid, tube, or many other shapes and sizes.

Alborosie2 years ago

8 0

Answer:

khansamueen has already provided an answer. This is an additional note on the many structures carbon may form, called allotropes.

Explanation: Carbon can take on many forms. They are all based only on the carbon atom, and are thus called alloptropes. One of these is diamond, "a" in the attached image.

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A cylinder with 10 mL of coconut oil has a mass of 9.3 g. What is the density of the oil? Round to the nearest hundredth.

lakkis [162]

Answer:

0.93 grams per milliliter

Explanation:

Density is the division of mass by volume. It is how "hefty" an object is; for example, wood isn't very "hefty" but metal is. The density is measured in mass over volume, so it is 9.3/10. After applying units, it is 0.93 grams per milliliter.

6 0

3 years ago

How many molecules of nitrogen gas are found in 0. 045 L of nitrogen gas at STP?

stepan [7]

Answer:

See below

Explanation:

.045 liter / 22.4 l / mole * 6.022 x 10^23 molecules/mole * 2 atoms/molecule =

( * 2 becuase nitrogen gas is diatomic)

7 0

1 year ago

Calculate the standard enthalpy change for the reaction at 25 ∘ 25 ∘ C. Standard enthalpy of formation values can be found in th

WINSTONCH [101]

<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$Mg(OH)_2(s)+2HCl(g)\rightarrow MgCl_2(s)+2H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]$

We are given:

$\Delta H_f_{(Mg(OH)_2(s))}=-924.5kJ/mol\\\Delta H_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H_f_{(MgCl_2(s))}=-641.8kJ/mol\\\Delta H_f_{(H_2O(g))}=-241.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ$

Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ

6 0

3 years ago

The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2

Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K = 6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0

3 years ago

10. Which term is same for one mole of O2 and one mole of ethane C2H6

melamori03 [73]

Answer:

b) Number of molecules

Explanation:

3 0

3 years ago