The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663
Answer:
Si hay algo que sucedió entre ustedes raro o un comportamiento que tuviste con esa persona o que te vio hacer asi como le pudieron decir algo de ti que no le gustara a tu amigo hace que se comporte raro contigo
Explanation:
According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate).
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71
Answer:
- <u>1. Equation: 2x + 3 = 9x - 11</u>
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- <u>2. Each row has 2 chairs</u>
Explanation:
The variable x represents the number of chairs in each row.
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<u>1. She can form 2 rows of a given length with 3 chairs left over.</u>
Thus, she has:
number of rows number of chairs in chairs number of chairs
each row left over she has
2 x 3 2x + 3
<u>2. She can form 9 rows of the same length if she gets 11 more chairs.</u>
That means that she is short in 11 chairs to have 9x chairs, or that she has 11 less chairs than 9x chairs. Then she has:
<u>3. Equation:</u>
Then, number of chairs she has is 2x + 3 and, also, 9x - 11, which allows to set the equation:
<u>4. Solve the equation:</u>
Therefore, each row has 2 chairs, and she has 2x + 3 = 4 + 3 = 7 chairs.
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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