Answer:
no it should not that place is historical and so they should make where u can visit but protect it as a historical landmark
When using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration. Option D
<h3>What are ion-selective electrodes?</h3>
Analytical chemistry is a science that deal with the measurement and detection of the accurate amount of a substance. Analytical chemistry plays a large role in environmental management as it helps in the determination of the levels of contaminants in a sample.
An ion selective electrode is used in analytical chemistry to measure the amount of a target ion by converting its activity into a measurable electrical signal.
Hence, when using ion-selective electrodes, to compensate for a complex or unknown matrix, the standard addition method can be used to determine the analyte concentration.
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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.
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Answer:
c. F1-
Explanation:
In this chemical reaction the expression is:
HF + NaF → NaHF2
The ion that always keep the negative charge is the fluorine with a -1, if in this mixture there is more positive ions (H1+) the negative ion (F1-) will join with them.
Remember that also the Cl1- will be free, but the fluorine is more reactive than the fluorine.
Answer is: mass of barium sulfate is 0.668 grams.
Chemical reaction: BaCl₂ + Na₂SO₄ → BaSO₄<span> + 2NaCl.
V(</span>BaCl₂) = 25.34 mL ÷ 1000 mL/L = 0.02534 L.
c(BaCl₂) = 0.113 mol/L.
n(BaCl₂) = V(BaCl₂) · c(BaCl₂).
n(BaCl₂) = 0.02534 L · 0.113 mol/L.
n(BaCl₂) = 0.00286 mol.
From chemical reaction: n(BaCl₂) : n(BaSO₄) = 1 : 1.
n(BaSO₄) = 0.00286 mol.
m(BaSO₄) = n(BaSO₄) · M(BaSO₄).
m(BaSO₄) = 0.00286 mol · 233.4 g/mol.
m(BaSO₄) = 0.668 g.
