Answer is: c. CuSO4 + 2NaOH yields Cu(OH)2 + Na2SO4.
Copper atom has oxidation number +2, sulfur atom oxidation number +6, oxygen has oxidation number-2, sodium has oxidation number +1 and hydrogen has +1 on both side of chemical reaction, so elements did not change their oxidation numbers.
Answer:
Hello There!!
Explanation:
The temperature stays the same when a solid is melting or a liquid is boiling (changing state) during a change of state, even though heat energy is being absorbed.
hope this helps,have a great day!!
~Pinky~
Answer:
THE MOLARITY IS 2.22 MOL/DM3
Explanation:
The solution formed was as a result of dissolving 37.5 g of Na2S in 217 g of water
Relative molecular mass of Na2S = ( 23* 2 + 32) = 78 g/mol
Molarity in g/dm3 is the amount of the substance dissolved in 1000 g or 1 L of the solvent. So we have;
37.5 g of Na2S = 217 g of water
( 37.5 * 1000 / 217 ) g = 1000 g of water
So, 172.81 g/dm3 of the solution
So therefore, molarity in mol/dm3 = mol in g/dm3 / molar mass
Molarity = 172.81 g/dm3 / 78 g/mol
Molarity = 2.22 mol/dm3
The molarity of the solution is 2.22 mol/dm3
Answer:
I wrote this answer to get points lol
Explanation:
Answer:
pH = 12.33
Explanation:
Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.
The titration reaction is
HA + KOH ---------------------------- A⁻ + H₂O + K⁺
number of moles of HA : 118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA
number of moles of OH : 115.4 mL/1000ml/L x 0.400 mol/L = 0.046 mol A⁻
therefore the weak acid will be completely consumed and what we have is the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.
n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH
pOH = - log (KOH)
M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M
pOH = - log (0.0021) = 1.66
pH = 14 - 1.96 = 12.33
Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.
