Answer : The volume of 6M NaOH stock solution is, 16.7 mL
Explanation :
To calculate the volume of NaOH stock solution, we use the equation given by neutralization reaction:
where,
are the molarity and volume of NaOH stock solution.
are the molarity and volume of NaOH.
We are given:
Putting values in above equation, we get:
Thus, the volume of 6M NaOH stock solution is, 16.7 mL
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:
4.43 g
Explanation:
The reaction between sodium chloride and flourine gas is given as;
NaCl + F2 --> NaF + Cl2
From the stochiometry of the equation;
1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2
Mass of 1 mol of F2 = 38g
Mass of 1 mol of sodium flouride, NaF = 42g
This means 38g of flourine reacted with NaCl to form 42g of NaF
xg of F2 would form 4.9g of NaF
38 = 42
x = 4.9
x = 4.9 * 38 / 42
x = 4.43 g
Answer:
Explanation:
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In this case, considering the given chemical reaction:
Thus, by applying the law of rate proportions, we can write:
Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:
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Answer:
43.05 moles of Al needed to react with 28.7 moles of FeO.
Explanation:
Given data:
Moles of FeO = 28.7 mol
Moles of Al needed to react with FeO = ?
Solution:
Chemical equation:
2Al + 3FeO → 3Fe + Al₂O₃
Now we will compare the moles of Al with FeO.
FeO : Al
2 : 3
28.7 : 3/2×28.7 = 43.05 mol
Thus 43.05 moles of Al needed to react with 28.7 moles of FeO.