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3241004551 [841]
4 years ago
7

What is the product of a force and the time during which the force acts

Chemistry
1 answer:
ELEN [110]4 years ago
7 0

Impulse is the product of a force and the time during which that force acts on a body.

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In performing this week's bromination reaction, if you were to start with 126 mg of acetanilide (135.17 g/mol), calculate the th
Artemon [7]

Answer:

Theoretical yield of C8H8BrNO:

In moles

0.000945

In grams

0.204

Explanation:

Theoretical yield of a reaction is defined as the quantity of the product obtained from the complete conversion of a limiting reactant in a chemical reaction. Theoretical yield can be expressed as grams or moles.

Equation of reaction:

C8H9NO + Br2 --> C8H8BrNO + HBr

Since C8H9NO is the limiting reagent, 1 mole of C8H9NO reacted to form 1 mole of C8H8BrNO

Mass of C8H9NO = 129 mg

= 0.129 g.

Molar mass of C8H9NO = 135.17 g/mol.

Number of moles of C8H9NO = mass/molar mass.

= 0.129/135.17

= 0.00095 moles of C8H9NO

Since 1 mole of C8H9NO yielded 1 mole of C8H8BrNO

Therefore, 0.000954 moles of C8H8BrNO

Theoretical yield (in grams) = molar mass * number of moles

= 214.06 * 0.00095

= 0.204 of C8H8BrNO

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3 years ago
Which of these is a base? a-vinegar b-ammonia c-HCl d- HNO3
Lady_Fox [76]
B- ammonia. ammonia has a pH level of about 11, and anything higher than 7  is more basic and anything less than 7 is more acidic. 7 is considered neutral.
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A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and
Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

    The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = 99.7^oC

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

1atm=760mmHg

or,

1mmHg=\frac{1}{760}atm

As, 1mmHg=\frac{1}{760}atm

So, 752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

K=273+^oC

As, K=273+^oC

So, K=273+99.7=372.7

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor  = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}

\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole

Thus, the mass of one mole of vapor is 32.0 g/mole

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No. The only thing that changed was the looks of the gasoline, not the chemical components.
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