That would be the first option Ca(OH)2 + H2SO4 → CaSO4 + 2H2O.
The Ca replaces the H2 in H2SO4, and the H2 replaces the Ca is Ca(OH)2.
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
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Answer:
a mixture of two these
Explanation:
The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.
n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.
Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.
Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.
Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.
Answer:
4.) 9, 1, and 4 5.) 4, 1, and 4
Explanation:
I am not quite sure about this because I cannot remember if the coefficient (the number before the elements) is applied to every element in the compound. If it is then your number of atoms are as follows: CORRECTION: you do not have to apply the coefficient to every element only the one that is after it. So when you back and fix the error your number of atoms will be as follows:
number 4
H: 9
P: 1
O: 4
number 5:
H: 4
S: 1
O: 4
you can calculate the number of atoms present in this compound by multiplying the coefficient and the subscripts of each atom.
hope this helped you :)
