B. The sand increases friction by increasing roughness.
Answer:
Percent yield = 84.5 %
Explanation:
Given data:
Mass of methanol = 229 g
Actual yield of water = 219 g
Percent yield of water = ?
Solution:
Chemical equation:
2CH₃OH + 3O₂ → 2CO₂ + 4H₂O
Number of moles of methanol:
Number of moles = mass/ molar mass
Number of moles = 229 g/ 32 g/mol
Number of moles = 7.2 mol
Now we will compare the moles of water with methanol.
CH₃OH : H₂O
2 : 4
7.2 : 4/2×7.2 = 14.4 mol
Mass of water:
Mass = number of moles × molae mass
Mass = 14.4 mol × 18 g/mol
Mass = 259.2 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 219 g / 259.2 g × 100
Percent yield = 84.5 %
Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
Explanation:
when one reactants is in excess, there will always be some left over. The other reactants becomes limiting factor and controls show much of each product is produced.while using excess percentage yields this is at the expense of atom economy.
Explanation:
The solution of the lactic acd and sodium lactate is referred to as a buffer solution.
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. In this case, the weak acid is the lactic acid and the conjugate base is the sodium lactate.
Buffer solutions are generally known to resist change in pH values.
When a strong base (in this case, NaOH) is added to the buffer, the lactic acid will give up its H+ in order to transform the base (OH-) into water (H2O) and the conjugate base, so we have:
HA + OH- → A- + H2O.
Since the added OH- is consumed by this reaction, the pH will change only slightly.
The NaOH reacts with the weak acid present in the buffer sollution.
