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Lelu [443]
3 years ago
8

A 25 mL syringe is used to measure both 10mL of a solution and 20mL of a solution. Which measurement, if any, has the biggest pe

rcentage error
Chemistry
1 answer:
Minchanka [31]3 years ago
5 0

All measuring devices have some given error in their measurements, such that the percent error is given by <u>100% times the quotient between the error and the measure itself.</u>

From this, we will see that the <u>10mL measurement has the largest percentage error.</u>

Here we have a 25 mL syringe, and we want to use it to measure both 10mL of a solution and 20 mL of a solution.

We do not to do any math to know that the smaller quantity will have a bigger percentage error.

Why? Well, because the syringe is the same in both cases, so the numerator in the fraction that gives the percentage error will be the same for both measures.

So the only thing that defines the percentage error will be the measure itself that goes in the denominator. Thus, <u>having a smaller measure means that the denominator is smaller</u>, so the fraction is larger,<u> thus the percentage error is larger.</u>

For example, just to also show some numbers, if the error of the syringe is 0.5 mL, the percentage error for the 10 mL measure is:

p₁ = 100%*(0.5 mL)/(10 mL) = 5%

While for the 20 mL measure the percentage error is

p₂ = 100%*(0.5mL/20mL) = 2.5%

If you want to learn more, you can read:

brainly.com/question/4170313

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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 %

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