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Leno4ka [110]
3 years ago
7

Aspirin can be made in the laboratory by reacting acetic anhydride (c4h6o3) with salicylic acid (c7h6o3) to form aspirin (c9h8o4

) and acetic acid (c2h4o2). the balanced equation is: c4h6o3+c7h6o3→c9h8o4+c2h4o2.in a laboratory synthesis, a student begins with 5.00 ml of acetic anhydride (density = 1.08 g / ml) and 2.08 g of salicylic acid. once the reaction is complete, the student collects 2.57 g of aspirin.
Chemistry
1 answer:
Temka [501]3 years ago
7 0
V(C₄H₆O₃) = 5.00 mL.
d(C₄H₆O₃) = 1.08 g/mL.
m(C₄H₆O₃) = V(C₄H₆O₃) · d(C₄H₆O₃).
m(C₄H₆O₃) = 5.00 mL · 1.08 g/mL.
m(C₄H₆O₃) = 5.4 g.
n(C₄H₆O₃) = m(C₄H₆O₃) ÷ M(C₄H₆O₃).
n(C₄H₆O₃) = 5.4 g ÷ 102 g/mol.
n(C₄H₆O₃) = 0.0529 mol.
n(C₇H₆O₃) = 2.08 g ÷ 138.1 g/mol.
n(C₇H₆O₃) = 0.015 mol; limiting reactant.
From chemical reaction: n(C₄H₆O₃) : n(C₉H₈O₄) = 1 : 1.
n(C₉H₈O₄) = 0.015 mol.
m(C₉H₈O₄) = 0.015 mol · 180.16 g/mol.
m(C₉H₈O₄) = 2.71 g; theoretical yield.
percent yield od aspirine = 2.57 g ÷ 2.71 g · 100% = 94.83%.

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8 0
3 years ago
3. 2.15 x 3.1 x 100 =
jeka94

Answer:

666.5

Explanation:

Multiply 2.15 and 3.1 to get 6.665.

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3 0
3 years ago
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Lead has a density of 11.3 g/mL. If you have 105 grams of lead,
Elza [17]

Answer:

<h2>Volume = 9.29 mL</h2>

Explanation:

Density of a substance can be found by using the formula

Density( \rho) =  \frac{mass}{volume}

From the question

Density = 11.3 g/mL

mass = 105 g

Substitute the values into the above formula and solve for the volume

That's

11.3 =  \frac{105}{v}

Cross multiply

11.3v = 105

Divide both sides by 11.3

v =  \frac{105}{11.3}

v = 9.29203

We have the final answer as

<h3>Volume = 9.29 mL</h3>

Hope this helps you

6 0
3 years ago
Which of the following is NOT an example of a Heterogeneous Mixture?
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5 0
2 years ago
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The specific heat of copper is 0.093 cal/g0C. Calculate the temperature change that occurs if 28 g of copper at 25 0C absorbs 58
Umnica [9.8K]

Answer:

22.27 °C = ΔT

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass = 28 g

heat absorbed = 58 cal

specific heat of copper =  0.093 cal/g .°C

temperature change =ΔT= ?

Solution:

Q = m × c × ΔT

58 cal = 28 g × 0.093 cal /g.°C × ΔT

58 cal = 2.604 cal.°C × ΔT

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22.27 °C = ΔT

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