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IgorC [24]
3 years ago
14

Nitrous oxide (N2O) is used as an anesthetic (laughing gas) and in aerosol cans to produce whipped cream. It is also a potent gr

eenhouse gas. The decomposition of N2O is first order. It decomposes slowly to N2 and O2:
2 N2O(g) --> 2 N2(g) + O2(g)

How many half-lives will it take for the concentration of the N20 to reach 3.13% of its original concentration?
Chemistry
1 answer:
scZoUnD [109]3 years ago
3 0

Answer:

five half lives

Explanation:

Half-life is the time required for a quantity to reduce to half of its initial value.

How many half lives it would take to reach 3.13% form 100% of it's initial concentration:

100% - 50% : First Half life

50% - 25%: Second Half life

25% - 12.5%: Third Half life

12.5% - 6.25%: Fourth Half life

6.25% - 3.125%: Fifth Half life

This means it would take five half lives to get to 3.125% (≈ 3.13%) of it's original concentration.

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An unstable nucleus which starts a decay process is called the _____.
xeze [42]
It Is Called The Parent Nuclide

8 0
3 years ago
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Identify the substance that conducts electricity. question 19 options: 1) nacl dissolved in water 2) solid nacl 3) solid sugar 4
Delicious77 [7]

The substance that conducts electricity is NaCl dissolved in water.

So, option A is correct one.

When the sodium chloride dissolve in water , the sodium atoms and chlorine atoms separates under the presence of water molecules and exist as sodium cation and chloride anion . Now , they are free to move around in the water as positively and negatively charged ions . This separation of charge allows the solution to conduct electricity.

The solid NaCl and solid sugar does not conduct electricity because it is not dissolve in water . Similarly, sugar dissolved in water does not conduct electricity .

to learn more about conduct electricity.

brainly.com/question/1458220

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4 0
1 year ago
The mass defect for the formation of lithium-6 is 0.0343 g/mol. The binding energy for lithium-6 nuclei is ________ kJ/mol. Ente
Sholpan [36]

<u>Answer:</u> The binding energy for lithium-6 nuclei is 3.09 E+11

<u>Explanation:</u>

Binding energy is defined as the energy which holds the nucleus together. It is basically the product of mass defect and the square of the speed of light.

This energy is calculated by using Einstein's equation, which is:

E=\Delta mc^2

where,

E = Binding energy of the atom

Delta m = Mass defect = 0.0343g/mol = 0.0343\times 10^{-3}kg/mol     (Conversion factor: 1kg=10^3g )

c = speed of light = 3\times 10^8m/s

Putting values in above equation, we get:

E=0.0343\times 10^{-3}kg/mol\times (3\times 10^8m/s)^2

E=3.09\times 10^{14}J/mol=3.09\times 10^{11}kJ/mol     (Conversion factor: 1kJ=10^3J )

Hence, the binding energy for lithium-6 nuclei is 3.09 E+11

7 0
3 years ago
How will the addition of a catalyst affect the reaction rate of the chemical reaction?
Rina8888 [55]

Answer:

It will cause the reaction rate to increase.

3 0
3 years ago
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Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.
nikdorinn [45]

Answer:

The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ

Explanation:

From Arrhenius equation showing the temperature dependence of reaction rates.

K = Ae^{\frac{Ea}{RT} } where

k = rate constant

A = Frequency or pre-exponential factor

Ea   =       energy of activation

R = The universal gas constant

T = Kelvin absolute temperature

we have

f = e^{\frac{Ea}{RT} }

Where

f = fraction of collision with energy higher than the activation energy

Ea = activation energy = 10.0kJ = 10000J

R = universal gas constant = 8.31 J/mol.K

T = Absolute temperature in Kelvin = 400K

In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature  

Plugging in the values into the equation relating f to activation energy we get

f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} } or f = e^{3.01} = 20.22 moles of argon have an energy of 10.0 kJ or greater

5 0
3 years ago
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