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Stella [2.4K]
2 years ago
6

Give an example of role of luck in bottenheims research

Chemistry
2 answers:
sattari [20]2 years ago
4 0

"You may find what you seek, or you might not" Is an excellent quote that might help.

Veronika [31]2 years ago
3 0

"You may find what you seek, or you might not" Is an excellent quote that might help.

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Which has not been suggested as a reasonably practical way to store large amounts of hydrogen in relatively small spaces for its
Yakvenalex [24]

Answer: A. Liquefy hydrogen under pressure and store it much as we do with liquefied natural gas today.

Explanation:

Current Hydrogen storage methods fall into one of two technologies;

  1. <em>physical storage</em> where compressed hydrogen gas is stored under pressure or as a liquid; and
  2. <em>chemical storage</em>, where the hydrogen is bonded with another material to form a hydride and released through a chemical reaction.

Physical storage solutions are commonly used technologies but are problematic when looking at using hydrogen to fuel vehicles. Compressed hydrogen gas needs to be stored under high pressure and  requires large and heavy tanks. Also, liquid hydrogen boils at -253°C (-423°F) so it needs to be stored cryogenically with heavy insulation and actually contains less hydrogen compared with the same volume of gasoline.  

Chemical storage methods allow hydrogen to be stored at much lower pressures and offer high storage performance due to the strong binding of hydrogen and the high storage densities. They also occupy relatively smaller spaces than either compressed hydrogen gas or liquified hydrogen. A large number of chemical storage systems are under investigation, which involve hydrolysis reactions, hydrogenation/dehydrogenation reactions, ammonia borane and other boron hydrides, ammonia, and alane etc.

Other practical storage methods being researched that focuses on storing hydrogen as a lightweight, compact energy carrier for mobile applications include;

  • Metal hydrides  e.g. LiH
  • Nanostructured metal hydrides
  • Non-metal hydrides
  • Carbohydrates
  • Synthesized hydrocarbons
  • Aluminum
  • Liquid organic hydrogen carriers (LOHC)
  • Encapsulation , e.t.c.
5 0
2 years ago
Be sure to answer all parts. Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectr
GenaCL600 [577]

Answer:

The possible structures are ketone and aldehyde.

Explanation:

Number of double bonds of the given compound is calculated using the below formula.

N_{db}=N_{c}+1-\frac{N_{H}+N_{Br}-N_{N}}{2}

N_{db}=Number of double bonds

N_{c} = Number of carbon atoms

N_{H} = Number of hydrogen atoms

N_{N} = Number of nitrogen atoms

The number of double bonds in the given formula - C_{4}H_{8}O

N_{db}= 4+1-\frac{8+0-0}{2}=1

The number of double bonds in the compound is one.

Therefore, probable structures is as follows.

(In attachment)

The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.

alkene compounds I and II shows signal less than 140 ppm.

Hence, the probable structures III and IV are given as follows.

The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.

Hence, the molecular formula of the compound C_{4}H_{8}O having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.

8 0
2 years ago
A graduated cylinder initially has 32.5 mL of water in it. After a 75.0 g piece of lead (Pb) is added to the graduated cylinder,
Mice21 [21]

Answer:

39.1-32.5 and you will find your answer it always like that, you subtract your starting point from your ending point

Explanation:

6 0
2 years ago
A geological sample is found to have a Pb-206/U-238 mass ratio of 0.337/1.00. Assuming there was no Pb-206 present when the samp
nikdorinn [45]

Answer:

2.1x10⁹ years

Explanation:

U-238 is a radioactive substance, which decays in radioactive particles. It means that this substance will lose mass, and will form another compound, the Pb-206.

The time need for a compound loses half of its mass is called half-life, and knowing the initial mass (mi) and the final mass (m) the number of half-lives passed (n) can be found by:

m = mi/2ⁿ

The mass of Pb-206 will be the mass that was lost by U-238, so it will be mi - m. Thus, the mass ration can be expressed as:

(mi-m)/m = 0.337/1

mi - m = 0.337m

mi = 1.337m

Substituing mi in the expression of half-life:

m = 1.337m/2ⁿ

2ⁿ = 1.337m/m

2ⁿ = 1.337

ln(2ⁿ) = ln(1.337)

n*ln(2) = ln(1.337)

n = ln(1.337)/ln2

n = 0.4190

The time passed (t), or the age of the sample, is the half-life time multiplied by n:

t = 4.5x10⁹ * 0.4190

t = 1.88x10⁹ ≅ 2.1x10⁹ years

5 0
2 years ago
Molar solubility of CuBr Ksp = 6.3 x 10-9
maksim [4K]
First, lets write the equation: CuBr--> Cu+2 + Br-
then, write the Ksp expression: Ksp= [Cu+2] [Br-]= 6.3 x 10-9
We let x be the concentration of Ksp that dissolved. Ksp= (x) (x)= 6.3 x 10-9.
Now we solve for x

X^2= 6.3x10-9
x= square root(6.3x10-9)
x= 7.94x10-5
6 0
2 years ago
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