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Naddik [55]
3 years ago
6

Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize t

etraphenylcyclopentadienone. You isolated 3.0 g of tetraphenylcyclopentadienone. What is the % yield
Chemistry
1 answer:
almond37 [142]3 years ago
3 0

Answer:

the % yield is 82%

Explanation:

Given the data in the question,

we know that;

Molar mass of benzil is 210.23 g·mol−1

Molar mass of dibenzyl ketone is 210.27 g·mol−1

Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1

Now,

2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole

2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole

3.0 g of tetraphenylcyclopentadienone = 3 / 384.5  = 0.0078 mole

Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone

percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%

= 0.82 × 100%

= 82%

Therefore, the % yield is 82%

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Answer:

D. The amount of heat required to increase the temperature of 1 g of a substance by 1 °C.

Explanation:

Specific heat is defined as the amount of heat needed to raise a unit of mass of a compound by one degree on the temperature scale.

The gram is constituted as a unit of mass, and the degree Celsius as a unit of temperature, therefore, the specific heat can be defined as the amount of heat required to increase the temperature of 1 g of a substance by 1 °C.

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3 years ago
What are some possible explanations you think a researcher might find to explain why the fish are washing up on the shore of Lak
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Answer:

the temperature or weather ,

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At room temperature (20°C} and pressure, the density of air is 1.189 g/L. An object will float in air if its density is less tha
Alekssandra [29.7K]

Explanation:

Density =\frac{Mass}[Volume}

Density of the air ,d= 1.189 g/L

(a) Density of the evacuated ball

Mass of the ball ,m = 0.12 g

Volume of the ball =V=560 cm^3=560 ml=0.560 L

D =\frac{0.12 g}{0.560 L}=0.214 g/L

D<d, teh evacuated ball will flaot in air.

(b) Density of the evacuated ball D = 0.214 g/L

Density of carbon dioxide gas = d_1=1.830 g/L

Mass of the carbon dioxide gas :

1.830 g/L\times 0.560 L=1.0248 g

Total density of filled ball with carbon dioxide gas:

\frac{0.12 g+1.0248 g}{0.560 L}==2.044 g/L

The ball filled with carbon dioxide will not float in the air because total density of filled ball is greater than the density of an air.

(c) Density of the evacuated ball D = 0.214 g/L

Density of hydrogen gas = d_2=0.0899 g/L

Mass of the hydrogen gas :

1.830 g/L\times 0.560 L=0.050344 g

Total density of filled ball with hydrogen gas:

\frac{0.12 g+0.050344 g}{0.560 L}==0.3041 g/L

The ball filled with hydrogen will float in the air because total density of filled ball is lessor than the density of an air.

(d) Density of the evacuated ball D = 0.214 g/L

Density of oxygen gas = d_3=1.330 g/L

Mass of the oxygen gas :

1.330 g/L\times 0.560 L=1.7448 g

Total density of filled ball with oxygen gas:

\frac{0.12 g+1.7448 g}{0.560 L}=1.5442 g/L

The ball filled with oxygen will not float in the air because total density of filled ball is greater than the density of an air.

(e) Density of the evacuated ball D = 0.214 g/L

Density of nitrogen gas = d_4=1.165 g/L

Mass of the nitrogen gas :

1.165 g/L\times 0.560 L=0.6524 g

Total density of filled ball with nitrogen gas:

\frac{0.12 g+0.6524 g}{0.560 L}==1.3792 g/L

The ball filled with nitrogen will not float in the air because total density of filled ball is greater than the density of an air.

f) Mass must be added to sink the ball = m

Density of ball > Density of the air ; to sink the ball.

\frac{0.12g +m}{0.560L}>1.189 g/L

m > 0.54584 g

For any case weight added to ball to make it sink in an air should be grater than the value of 0.54584 grams.

5 0
2 years ago
A 0.187 M weak acid solution has a pH of 3.99. Find Ka for the acid. Express your answer using two significant figures.
Anna35 [415]

Answer:

5.56 × 10⁻⁸

Explanation:

Step 1: Given data

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  • pH of the solution: 3.99

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -3.99 = 1.02 × 10⁻⁴ M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(1.02 \times 10^{-4})^{2} }{0.187} = 5.56 \times 10^{-8}

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10. How much protein one should obtain from diet<br> (1 gram, 3 gram,4gram)
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How much protein one should obtain from diet

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4 gram

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