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2 years ago

Tritium 3 1H decays to 3 2He by beta emission. Find the energy released in the process. Answer in units of keV.

Chemistry

1 answer:

-Dominant- [34]2 years ago

7 0

Answer:

The energy released in the decay process = 18.63 keV

Explanation:

To solve this question, we have to calculate the binding energy of each isotope and then take the difference.

The mass of Tritium = 3.016049 amu.

So,the binding energy of Tritium = 3.016049 *931.494 MeV

= 2809.43155 MeV.

The mass of Helium 3 = 3.016029 amu.

So, the binding energy of Helium 3 = 3.016029 * 931.494 MeV

= 2809.41292 MeV.

The difference between the binding energy of Tritium and the binding energy of Helium is: 32809.43155 - 2809.412 = 0.01863 MeV

1 MeV = 1000keV.

Thus, 0.01863 MeV = 0.01863*1000keV = 18.63 keV.

So, the energy released in the decay process = 18.63 keV.

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Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated

Luden [163]

Complete Question

The complete question is shown on the first uploaded image

Answer

As the valve is opened , the gas will flow into the empty container until the both containers have the same pressure

$\Delta H = 0\ , \Delta E = 0 , q= 0 , w= 0$

The driving force for this process is the increase in entropy this is because the movement of the internal energy of the gas into a larger volume, what this does is that it increases the amount of disorder(entropy).

Explanation

In order to obtain the parameter in the part B of the question we are first obtain the initial pressure, using the ideal gas equation

$P = \frac{nRT}{V}$

$P = \frac{(2.4mol)(0.0821\frac{1 atm}{K \cdot \ mol} )}{4.0L}$

$P =15 \ atm$

The next thing is to obtain the new pressure of the gas , using boyle's law

$P_1V_1 = P_2V_2$

$P_2 = \frac{P_1 V_1}{V_2}$

$P_2 = \frac{(15 \ atm)(4.0L)}{24.0 L}$

$P_2 = 2.5 \ atm$

Since the this process is isothermal , the change in heat is equal to zero

i.e q = 0 J

The workdone to move the gas to the other container is zero because the the pressure at this second container is zero due to the fact that it is a vacuum

i.e $w = -P_{external} \Delta V$

$=-(0 \ atm) (24.0 - 4.0L)$

$= 0L \cdot atm$

Since the change in heat is zero and the workdone is zero then the change in internal energy is equal to 0

This is because the change in internal energy is equal to a summation of change in heat and the workdone

i.e $\Delta E = q + w$

$= 0J$

Generally the change in enthalpy is mathematically represented as

$\Delta H = n C_p \Delta T$

Since the temperature is zero this means that the change in temperature is zero , substituting this value for change in temperature into the equation for change in enthalpy

$\Delta H = n C_p (0)$

$= 0J$

7 0

2 years ago

Which isolated gaseous atom has a total of five electrons occupying spherically shaped orbitals?

Andre45 [30]

Let's find the electron configuration of each atom

Atom atomic number electron configuration

Bo 5 1s2 2s2 2p
F 9 1s2 2s2 2p5
Na 11 1s2 s2 2p6 3s
K 19 1s2 2s2 2p6 3s2 3p6 4s

The spherically shaped orbitals are the s orbitals.

You can se in the table that I wrote above that sodium has 2 electrons at 1s, 2 electrons at 2s and 1 electron at 3s orbitals, that is a total of five electrons at s (spherically shaped orbtials).

So the answer is the option A) sodium.

3 0

2 years ago

Sodium thiosulfate (Na2S2O3), photographer’s

vova2212 [387]

3Na2S2O3 + AgBr ------>Na5[Ag(S2O3) 3] +NaBr

from equation 3 mol 1 mol

given x mol 0.10 mol

x= (3*0.10)/1=0.30 mol Na2S2O3

Answer: 0.30 mol Na2S2O3

3 0

2 years ago

Read 2 more answers

How many hydrogen bonds can CH2O make to water

VladimirAG [237]

Hydrogen bonds are not like covalent bonds. They are nowhere near as strong and you can't think of them in terms of a definite number like a valence. Polar molecules interact with each other and hydrogen bonds are an example of this where the interaction is especially strong. In your example you could represent it like this:

H2C=O---------H-OH 

But you should remember that the H2O molecule will be exchanging constantly with others in the solvation shell of the formaldehyde molecule and these in turn will be exchanging with other H2O molecules in the bulk solution. 

Formaldehyde in aqueous solution is in equilibrium with its hydrate. 

H2C=O + H2O <-----------------> H2C(OH)2

5 0

2 years ago

Which postulate of Dalton's theory is consistent with the following observation concerning the weights of reactants and products

hram777 [196]

Answer: This illustrates law of conservation of mass.

Explanation:

Dalton's theory is based on mainly two laws which are law of conservation of mass and law of constant proportion.

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the decomposition of calcium carbonate follows:

$CaCO_3\rightarrow CaO+CO_2$

We are given:

Mass of calcium carbonate = 100 grams

Mass of calcium oxide = 56 grams

Mass of carbon dioxide = 44 grams

Total mass on reactant side = 100 g

Total mass on product side = 56 + 44 = 100 g

As, the total mass on reactant side is equal to the total mass on product side.

Thus, this illustrates law of conservation of mass.

6 0

2 years ago