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3 years ago

¿que se necesita para escribir un informe de experimento? Plis

Chemistry

2 answers:

s2008m [1.1K]3 years ago

5 0

Answer:

Por favor, dale a la otra persona lo más inteligente. Gracias

Ludmilka [50]3 years ago

4 0

Answer:

Explanation:

Introducción. Dentro de la introducción, el investigador limita el estudio. Muestra los datos generales importantes y...

Hipótesis. La hipótesis consiste en una apreciación que el investigador hace en torno al problema. Se define como una...

Objetivos. Los objetivos de un estudio son los movimientos que se deben hacer para solucionar el

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Air is made of nitrogen, carbon dioxide, and small amounts of other gases. What is the pressure of O₂, total is 1 atm if PN₂ = 5

Stolb23 [73]

Answer:

The pressure of O₂ is 0.8 atm.

Explanation:

The pressure exerted by a particular gas in a mixture is known as its partial pressure. So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case:

PT=Pnitrogen + Pcarbon dioxide + Pother gases

Being:

Pnitrogen: 593.4 mmHg
Pcarbon dioxide: 3 mmHg
Pother gases: 7.1 mmHg

and replacing:

PT= 593.4 mmHg + 3 mmHg + 7.1 mmHg

you get:

PT= 603.5 mmHg

Being 760 mmHg= 1 atm, you get:

PT= 603.5 mmHg= 0.8 atm

<u><em>The pressure of O₂ is 0.8 atm.</em></u>

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Ehem I have thirty levels in minecraft and a full enchantment room set up what should I put it on? My diamond axe (which is my w

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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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21. _____ Which of the following entities has the greatest mass?

IrinaK [193]

I believe it’s D because phosphorus has the mass of 30.9.

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2 PUIS

avanturin [10]

Answer: C the study of living things

Explanation:

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¿que se necesita para escribir un informe de experimento? Plis ​

¿que se necesita para escribir un informe de experimento? Plis