The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol in 1 L solution
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L
therefore volume of acid required = 8 mL
Answer:
D
Explanation:
There are principally three states of matter. These are the solid, liquid and gaseous states. The gaseous state has the highest degree of disorderliness as gas particles can move randomly while the solid state has the highest level of compactness.
Hence, we need to be adequately fed with information as regards the phase change to know if entropy has decreased or increased.
A. is wrong
Evaporation is a change of state to the gaseous state meaning there is an increased entropy.
B. is wrong
Sublimation is a change of state which means a solid substance like iodine or naphthalene changes its state directly to the gaseous state. There is an increased entropy here too.
C is wrong
Melting of ice means going from ice block to liquid water. This is synonymous to going from the solid state to the liquid state which is an increased entropy
D is correct
Condensation involves going from the gaseous state to the liquid state. This means going from a less ordered state to a more ordered state. This is accompanied by an entropy decrease.
E is wrong
While there are some processes that increase entropy, we also have some process that decrease entropy.
Answer:
balanced equation mole ratio 5 2 mol NO/1 mol O2
10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2
20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO
actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2
Because the actual mole ratio of NO:O2 is larger than the balanced equation mole
ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.
Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO
0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO
Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2
Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N
Explanation:
