True or False: Cohesion is the attraction between particles of the same<br> substance

What is the density of ammonia (NH3) at 293 K and 0.913 atm?

Anastaziya [24]

Answer: The density of Ammonia is 0.648 g/l

Explanation:

Density = Mass/ Volume

Mass of one mole of Ammonia (NH3) = 17.031g

Volume =?

Using the ideal gas law we can determine the volume.

PV = nRT

P = 0.913 atm, V= ?, n = 1, R = 0.08206 L.atm/K, and T= 293K

Make V the subject of the formular, we then have;

V= nRT/ P = 1 mol x 0.08206 L.atm/ K.mol x 293 / 0.913 atm

V = 24.04358/ 0.913 = 26.3L

Having gotten the value of Volume in this question, we then go back to solve for density.

Density = Mass/ Volume

17.031g/ 26.3L = 0.64756 ≈ 0.648 g/l

7 0

3 years ago

Crude oil is ________. a. composed of less than ten different hydrocarbon molecules b. formed in a wide range of temperature and

Arisa [49]

Answer:

C

Explanation:

A - Crude oil is composed for hundreds of hydrocarbon, not less than ten.

B - Is formed in specific conditions of temperature and pressure

C - It's fractionated to form gasoline, lubricants, CH4, plastics and many other products made of hydrocarbon.

D - We have crude oil located more than 4000 yards below the surface in countries like Brazil

E - The crude oil is very thick and don't have an specific usage, so we need to refine it.

3 0

3 years ago

Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

gulaghasi [49]

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

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5 0

2 years ago

What do acids feel like

djverab [1.8K]

Answer:

In fact, some strong bases can burn the skin as badly as strong acids. Bases feel soapy or slippery because they react with acidic molecules in your skin called fatty acids. ... Like acids, bases change the colors of acid-base indicators, but the colors they produce are different. Bases turn litmus paper blue.

3 0

3 years ago

If the half-life of a radioactive substances is 590 million years and you have 40 atoms of it, how many half-lives will have pas

Nady [450]

Answer:

3

Explanation:

Applying,

$2^{n'}$ = R/R'............... Equation 1

Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.

From the question,

Given: R = 40 atoms, R' = 5 atoms

Substitute these values into equation 1

$2^{n'}$ = 40/5

$2^{n'}$ = 8

$2^{n'}$ = 2³

Equation the base,

n' = 3

3 0

3 years ago

True or False: Cohesion is the attraction between particles of the same substance