Answer:you observe the light?
Explanation:
Answer:
C. A series circuit has only one loop and a parallel circuit has two or more loops for the current to flow through
Explanation:
A circuit that are made of one loop is called series circuit. On the other hand, the parallel circuit has at least two loops. The circuit type has nothing to do with open or closed circuit.
If any part of the series circuit got cut, the current will stop flowing since there is only one loop. A parallel circuit has more loop so the circuit might still work even if a part of the circuit got cut.
The pH = 2.41
<h3>Further explanation</h3>
Given
5.0% by mass solution of acetic acid
the density of white vinegar is 1.007 g/cm3
Required
pH
Solution
Molarity of solution :
Ka for acetic acid = 1.8 x 10⁻⁵
[H⁺] for weak acid :
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Input the value :
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.839%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D0.00388%3D3.88%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~3.88%3D2.41)
Answer:
Like other alkali metals, rubidium metal reacts violently with water. As with potassium (which is slightly less reactive) and caesium (which is slightly more reactive), this reaction is usually vigorous enough to ignite the hydrogen gas it produces.
Explanation:
hope it helps
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:
brainly.com/question/24311846
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