Answer:
The metals that would be able to reduce copper ions in solution would be hydrogen, lead, tin, nickel, iron, zinc, aluminum, magnesium, sodium, calcium, potassium, silver, gold, copper and lithium 5.
Explanation:
The product for the following reaction are :
- 2 - butanone
- CH₃CH₂COCl
- CH₃CH₂CONHCH₃
T carboxylic acid is an organic compound. the functional group of carboxylic acid is carboxy. general formula is given as : R - COO⁻.
The reactions are given as follows :
1) the reduction of carboxylic acid into ketone with the organometallic compound is given as follows :
CH₃CH₂COOH + CH₃Li ------> CH₃CH₂COCH₃
2 butanone
2) the reaction of carboxylic acid with SOCl₂ is given as follows :
CH₃CH₂COOH + SOCl₂ -----> CH₃CH₂COCl
3) the reaction of carboxylic acid with CH₃NH₂ is given as follows
CH₃CH₂COOH + CH₃NH₂ -----> CH₃CH₂CONHCH₃
Thus, The product for the following reaction are :
- 2 - butanone
- CH₃CH₂COCl
- CH₃CH₂CONHCH₃
To learn more about carboxylic acid here
brainly.com/question/4721247
#SPJ4
Answer is: <span>volume of oxygen is 14.7 liters.
</span>Balanced chemical
reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.<span>
m(</span>C₃H₈-propane) = 5.53 g.
n(C₃H₈) = m(C₃H₈) ÷ M(C₃H₈).
n(C₃H₈) = 5.53 g ÷ 44.1 g/mol.
n(C₃H₈) = 0.125 mol.
From chemical reaction: n(C₃H₈) : n(O₂) = 1 : 5.
n(O₂) = 0.625 mol.
T = 25° = 298.15K.
p = 1.04 atm.
<span>R = 0.08206
L·atm/mol·K.
</span>Ideal gas law: p·V = n·R·T.
V(O₂) = n·R·T / p.
V(O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.
V(O₂) = 14.7 L.
Answer:
One of the most common neutralization reaction occurs whensodium hydroxide(NaOH) reacts withhydrochloric acid(HCl). The product formed are salt(NaCl) and water(H2O).
Explanation:
The volume of 0.15M NaOH that could be produced with only 5 grams of NaOH and water is 0.83L.
<h3>How to calculate volume?</h3>
The volume of a substance can be calculated using the following expression:
Molarity = number of moles ÷ volume
The number of moles of NaOH can be calculated by dividing the mass of NaOH by its molar mass.
Molar mass of NaOH = 40g/mol
n = 5/40
n = 0.125moles
Volume of NaOH = 0.125mol ÷ 0.15M
Volume = 0.83L
Therefore, the volume of 0.15M NaOH that could be produced with only 5 grams of NaOH and water is 0.83L.
Learn more about volume at: brainly.com/question/905400
