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3 years ago

If 65.0 g of aspirin (C₉H₈O₄) are produced from 79.8 g of C₇H₆O₃, what is the percent yield from the reaction below?

Chemistry

1 answer:

lyudmila [28]3 years ago

7 0

Answer:

f 65.0 g of aspirin (C₉H₈O₄) are produced from 79.8 g of C₇H₆O₃, what is the percent yield from the reaction below?

C₇H₆O₃ (s) + C₄H₆O₃ (s) → C₉H₈O₄ (s) + HC₂H₃O₂ (aq).

Explanation:

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A material has a volume of 63.0 cm3 and a mass of 28 grams. What is the density of the material in g/cm3 to the correct number

dimaraw [331]

Answer:

0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).

Explanation:

Knowing that:

d = m/V,

where, d is the density of the material (g/cm³).

m is the mass of the material (m = 28 g).

V is the volume of the material (V = 63.0 cm³).

∴ d = m/V = (28 g)/(63.0 cm³) = 0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).

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3 years ago

How do kinetic tiles work?

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4 years ago

7.5 moles of nitrogen gas (N2) is formed in the following reaction. How many grams

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Answer:

Mass = 255 g

Explanation:

Given data:

Number of moles of nitrogen = 7.5 mol

Mass of ammonia formed = ?

Solution:

Chemical equation:

3H₂ + N₂ → 2NH₃

Now we will compare the moles of nitrogen and ammonia.

N₂ : NH₃

1 : 2

7.5 : 2/1×7.5 = 15

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3 years ago

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2 years ago

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A lead mass is heated and placed in a foam cup calorimeter containing 40.0 mL of water at 17.0°C. The water reaches a temperatur

lbvjy [14]

Answer: 502 Joules

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 40.0 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{40.0mL}\\\\\text{Mass of water}=(1g/mL\times 40.0mL)=40.0g$

When metal is dipped in water, the amount of heat released by lead will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$q=m\times c\times \Delta T$

q = heat absorbed by water

$m$ = mass of water = 40.0 g

$T_{final}$ = final temperature of water = 20.0°C

$T_{initial$ = initial temperature of water = 17.0°C

$c$ = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$q=40.0\times 4.186\times (20.0-17.0)]$

$q=502J$

Hence, the joules of heat were re-leased by the lead is 502

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3 years ago