Answer: hydrolysis oxidation acidification
Explanation:
they break down, dissolve, or create new elements
Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
covalent bond is firmed between two atoms
Answer:
0.166M
Explanation:
In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:
<em>Moles H2SO4:</em>
0.650L * (0.430mol /L) = 0.2795moles H2SO4
<em>Moles KOH:</em>
0.600L * (0.240mol / L) = 0.144 moles KOH
Moles of sulfuric acid that reacts with 0.144 moles of KOH are:
0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.
And remain:
0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.
In 0.650L + 0.600L = 1.25L:
Molar concentration of sulfuric acid:
0.2075 moles of H2SO4 / 1.25L =
<h3>0.166M</h3>