Answer:
Precise but not accurate.
Explanation:
We can tell the performance of the balance is precise, because the repeated measurements give values close to one another.
However, the performance of the balance is not accurate, as the mean value of the repeated measurements (195.587) is not close to the value considered as true (in this case the standard calibration mass with a certified value of 200.002 g).
<u>We are given:</u>
M1 = 3 Molar V1 = 80 mL
M2 = x Molar V2 = 100 mL
<u>Finding the molarity:</u>
We know that:
M₁V₁ = M₂V₂
where V can be in any units
(3)(80) = (x)(100)
x = 240/100 [dividing both sides by 100]
x = 2.4 Molar
This is an acid-base reaction where HF is the acid and H2O is the base (it's amphoteric and can be an acid or a base). The products would then H3O+ (the conjugate acid) and F- (the conjugate base). Now, we can simply construct a reaction using the found products and reactants. This acid-base reaction would be HF + H2O <--> H3O+ + F-.
Hope this helps!
During the electrolysis of the molten lithium chloride, the Lithium ions (Li⁺) at the cathode undergoes reduction, and the electron configuration of lithium becomes 1s²2s¹.
<h3>What is electrolysis?</h3>
Electrolysis can be described as the process in which the electric current is passed through the chemical compound to break them. In this process, the atoms and ions are interchanged by the addition or removal of electrons.
The ions are allowed to move freely in this process. When an ionic compound is melted or dissolved in water then ions are produced which can move freely.
During the electrolysis of molten lithium chloride, the lithium ions reach the cathode and accept the electrons while chloride ions reach at anode and loss electrons to become chlorine gas.
At anode : 2 Cl⁻ → Cl₂ + 2e⁻
At cathode: 2 Li⁺ + 2e⁻ → Li
Learn more about electrolysis, here:
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Answer:
2.99 M
Explanation:
In order to solve this problem we need to keep in mind the definition of molarity:
- Molarity = moles of solute / liters of solution
In order to calculate the moles of solute, we <u>convert 125.6 g of NaF into moles</u> using its <em>molar mass</em>:
- 125.6 g NaF ÷ 42 g/mol = 2.99 mol NaF
As the volume is already given, we can proceed to <em>calculate the molarity</em>:
- Molarity = 2.99 mol / 1.00 L = 2.99 M