Answer:
16.0%.
Explanation:
Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.
V % of ethanol = (volume of ethanol / volume of the solution) x 100.
volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.
∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.
The correct answer should be that Potential energy increases, and kinetic energy increases, since they both increase as the temperature changes.
Answer:
The equilibrium shifts to the left, and the concentration of Ba2+(aq) decreases
Explanation:
Whenever a solution of an ionic substance comes into contact with another ionic compound with which it shares a common ion, the solubility of the ionic substance in solution decreases significantly.
In this case, both BaSO4 and Na2SO4 both possess the SO4^2- anion. Hence SO4^2- anion is the common ion. Given the equilibrium;
BaSO4(s) <—> Ba2+ (aq) + SO4 2- (aq), addition of Na2SO4 will decrease the solubility of BaSO4 due to the presence of a common SO4^2- anion compared to pure water.
This implies that the equilibrium will shift to the left, (more undissoctiated BaSO4) hence decreasing the Ba^2+(aq) concentration.
An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between at
Moles of potassium permanganate = 0.0008
<h3>Further explanation </h3>
Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range
Reaction
5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10CO2(g) + 8H2O(1)
The end point ⇒titrant and analyte moles equal
titrant : potassium permanganate-KMnO4
analyte : sodium oxalate - Na2C2O4
so moles of KMnO4 = moles of Na2C2O4
moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :
From equation, mol ratio Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :
