Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Answer:
The answer to your question is: Excess oxygen = 2.3 mol
Explanation:
Data
ZnS = 5 mol
O2 = 9.8 mol
Excess reactant = ?
Balanced reaction
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
MW ZnS = 65 + 32 = 97 x 2 = 194 g
MW O2 = 16 x 6 = 96 g
2 mol of ZnS ------------------- 3 mol O2
Ratio from the reaction = 3 mol O2/ 2 mol ZnS
= 1.5
Ratio from the quantities in the experiment = 9.8 mol O2 / 5 mol of ZnS
= 1.96
Excess reactant = Oxygen because the ratio increases
2 mol of ZnS ------------------- 3 mol O2
5 mol of ZnS ------------------- x
x = (5 x 3) / 2
x = 7.5 mol of O2
Excess Oxygen = 9.8 mol - 7.5 mol
Excess oxygen = 2.3 mol
Answer is: <span>the true volume of the flask is 10.2163 grams.
d</span>(H₂O) = 0.9981 g/mL; density of water at 20°C.
V(H₂O) = 10 mL; volume of water.
m(H₂O) = d(H₂O) · V(H₂O).
m(H₂O) = 0.9981 g/mL · 10 mL.
m(H₂O) = 9.9981 g.
m(empty flask) = m(filled flask) - m(H₂O).
m(empty flask) = 20.2144 g - 9.9981 g.
m(empty flask) = 10.2163 g.
Answer:
14.9075 g, 28.67%, 0.11%
Explanation:
The mean concentration of calcium = summation x / frequency
= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g
Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164
b) percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%
c) relative standard deviation = standard deviation / mean × 100 = 0.0164 / 14.9075 × 100 = 0.11%
d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.
The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.
Reactions of Ethyl-3-pentenoate with all given reagents are given below.
Reaction with H₂ / Pd:
The non-polar double bond present in Ethyl-3-pentenoate is reduced to saturated chain. This reagent can not reduce the carbonyl group.
Reaction with NaBH₄: Sodium Borohydride is a weak reducing agent at compared to LiAlH₄. It can only reduce aldehydes and Ketones to corresponding alcohols.
Reaction with LiAlH₄: Lithium Aluminium hydride is a strong reducing agent. It can reduce all types of carbonyl compounds to corresponding alcohols, But, it can not reduce non-polar double bonds like alkenes and alkynes.
Result: The correct answer is
Option-A (Highlighted RED below).
