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Genrish500 [490]
3 years ago
14

How many quarts are in 0.15 ML

Chemistry
1 answer:
olganol [36]3 years ago
7 0

Answer: divide by 946.353

= 1.58 x 10^-4 or 0.000159

Explanation:

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PLEASE HELP!!!!!!
pogonyaev

Answer:

m_{B}^{theoretical}=0.365gB

Y=87.1\%

Explanation:

Hello there!

In this case, since the reaction (A->B) have an initial amount of pure 4-aminobenzoic acid, the first step to compute the theoretical yield is to solve the following stoichiometric setup:

m_{B}^{theoretical}=0.303gA*\frac{1molA}{137.14gA}*\frac{1molB}{1molA}*\frac{165.19 gB}{1molB}\\\\   m_{B}^{theoretical}=0.365gB

Whereas A stands for 4-aminobenzoic acid and B for the benzocaine. Moreover, we compute the percent yield by dividing the actual yield (0.318 g) by the theoretical one (0.365 g):

Y=\frac{0.318g}{0.365g} *100\%\\\\Y=87.1\%

Best regards!

4 0
2 years ago
UGRENT! Please help showing all work
agasfer [191]

Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

8 0
3 years ago
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