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koban [17]
3 years ago
13

4. When you completed your experiment, you measured that 0.99 L of CO2 was released,

Chemistry
1 answer:
sergey [27]3 years ago
7 0

Answer:

Sodium Carbonate was the unknown powder

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What is occurring in the image below?
DiKsa [7]

The addictive quality is a reduction in vapour pressure.

The vapour pressure at a liquid's normal boiling point is the same as the ordinary atmospheric pressure, which is 1 atmosphere, 760 Torr, 101.325 kPa, or 14.69595 psi.

The pressure that results from liquids evaporating is known as vapour pressure. Surface area, intermolecular forces, and temperature are three often occurring variables that affect vapour press.

lower vapour pressure

raising the boiling point

Low-temperature depression

Osmotic force

They are all dependent on the solute; when you increase the solute, the colligative property and the ratio you added may change.

The Van't Hoff Factor is another option to examine (i). the number of dissolved ions. The colligative property will be further altered if the solute is ionic.

Learn more about Vapor pressure here brainly.com/question/14949391

SPJ1

3 0
1 year ago
Which element is likely to have the highest thermal conductivity
AlexFokin [52]
Gallium, due to its metal properties, which tend to conduct electricity and heat quite well.
3 0
3 years ago
What was a direct result of the Bay of Pigs invasion
Dafna1 [17]

2) Cold War tensions increased.

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3 0
3 years ago
Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/
Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of C_2H_4 and H_2 = 25.0 atm

Temperature of C_2H_4 and H_2 = 250^oC=273+250=523K

Volume of C_2H_4 = 1050 L per min

Volume of H_2 = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of C_2H_6 = 30 g/mole

First we have to calculate the moles of C_2H_4 and H_2 by using ideal gas equation.

For C_2H_4 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}

n=611.34moles

For H_2 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}

n=902.46moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_2H_4+H_2\rightarrow C_2H_6

From the balanced reaction we conclude that

As, 1 mole of C_2H_4 react with 1 mole of H_2

So, 611.34 mole of C_2H_4 react with 611.34 mole of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and C_2H_4 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of C_2H_6.

As, 1 mole of C_2H_4 react to give 1 mole of C_2H_6

As, 611.34 mole of C_2H_4 react to give 611.34 mole of C_2H_6

Now we have to calculate the mass of C_2H_6.

\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6

\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g

The theoretical yield of C_2H_6 = 18340.2 g

The actual yield of C_2H_6 = 14.0 kg = 14000 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of C_2H_6

\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%

Therefore, the percent yield of the reaction is, 76.34 %

5 0
3 years ago
HELP ASP
Kipish [7]

True becuase the substance changed into another substance which is a example of a chemical reaction.

3 0
3 years ago
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