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harina [27]
3 years ago
8

The average atomic mass of all naturally occurring lithium atoms is 6.941 amu. the two isotopes of lithium are lithium-6 and lit

hium-7. are these isotopes equally common? if not, which is more plentiful in nature, and how do you know?
Chemistry
1 answer:
Gnom [1K]3 years ago
6 0
No, the two isotopes of lithium-6 and lithium-7 are not equally common.  
The more plentiful isotope would be lithium-7. 
This can be easily demonstrated by assuming that both isotopes were equally common. If that were the case, the average atomic mass would be (6 + 7)/2 = 6.5 amu. Now compare that value if they were both equal to the actual value found in nature. The value found in nature is 6.941 amu which is heavier than the 6.5 amu that would happen if they were equally common. Since the natural value is heavier, that means that there has to be more of the heavier isotope than there is of the lighter one. Therefore lithium-7 is more common than lithium-6.
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The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+X_2^4\alpha+Y_{-1}^0e

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

_{92}^{235}\textrm{U}\rightarrow _{82}^{207}\textrm{Pb}+7_2^4\alpha+4_{-1}^0e

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

3 0
3 years ago
Reaction is
-Dominant- [34]
According  to the task, you  are proveded with patial pressure of CO2 and graphite, and here is complete solution for the task :
At first you have to find n1 =moles of CO2 and n2 which are moles of C 
<span>The you go :
</span>CO2(g) + C(s) \ \textless \ =\ \textgreater \  2CO(g)

n1 n2 0
-x -x +2x
n1-x n2-x 2x Kp=P^2 (CO)/Pco2
After that you have to use the formula PV=nRT
 Kp=(2x)^2/n1-x
Then you have to solve x, and for that you have to use <span>RT/V
And to find total values:</span>P(total)=n(total)*R*T/V
I am absolutely sure that this would be helpful for you.
7 0
3 years ago
Consider the letters in the word Chemistry. Use them to make as many word compounds as is possible with 9 elements. How is an el
dimaraw [331]
<span>The difference between an element and a compound is that an element is a substance made of same type of atoms, whereas a compound is made of different elements in definite proportions. Examples of elements include iron, copper, hydrogen and oxygen. Examples of compounds include water (H2O) and salt (Sodium Chloride - NaCl) </span>
6 0
3 years ago
What type of energy results from the
jeka94

Answer:

C fghhtrsfeagutyrwraqedf

3 0
3 years ago
A helium-filled balloon at 310.0 K and 1 atm, contains 0.05 g He, and has a volume of 1.21 L. It is placed in a freezer (T = 235
trapecia [35]

Answer : The value of \Delta E of the gas is 2.79 Joules.

Explanation :

First we have to calculate the moles of helium.

\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}

Molar mass of helium = 4 g/mole

\text{Moles of helium}=\frac{0.05g}{4g/mole}=0.0125mole

Now we have to calculate the heat.

Formula used :

q=nc_p\Delta T\\\\q=nc_p(T_2-T_1)

where,

q = heat

n = number of moles of helium gas = 0.0125 mole

c_p = specific heat of helium = 20.8 J/mol.K

T_1 = initial temperature = 310.0 K

T_2 = final temperature = 235.0 K

Now put all the given values in the above formula, we get:

q=nc_p(T_2-T_1)

q=(0.0125mole)\times (20.8J/mol.K)\times (235.0-310.0)K

q=-19.5J

Now we have top calculate the work done.

Formula used :

w=-p\Delta V\\\\w=-p(V_2-V_1)

where,

w = work done

p = pressure of the gas = 1 atm

V_1 = initial volume = 1.21 L

V_2 = final volume = 0.99 L

Now put all the given values in the above formula, we get:

w=-p(V_2-V_1)

w=-(1atm)\times (0.99-1.21)L

w=0.22L.artm=0.22\times 101.3J=22.29J

conversion used : (1 L.atm = 101.3 J)

Now we  have to calculate the value of \Delta E of the gas.

\Delta E=q+w

\Delta E=(-19.5J)+22.29J

\Delta E=2.79J

Therefore, the value of \Delta E of the gas is 2.79 Joules.

3 0
3 years ago
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