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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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How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0
3 years ago
What is a semiconductor?
makkiz [27]
It is c: a conductor that operates only at low temperatures
8 0
3 years ago
Determine the final temperature of sample with a specific heat of 1.1 J/g°C and a mass of 385 g if it starts out at a temperatur
Assoli18 [71]

Answer:

T2 =21.52°C

Explanation:

Given data:

Specific heat capacity of sample = 1.1 J/g.°C

Mass of sample = 385 g

Initial temperature = 19.5°C

Heat absorbed = 885 J

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

885J = 385 g× 1.1 J/g.°C×(T2 - 19.5°C )

885 J = 423.5 J/°C× (T2 - 19.5°C )

885 J / 423.5 J/°C = (T2 - 19.5°C )

2.02°C = (T2 - 19.5°C )

T2 = 2.02°C + 19.5°C

T2 =21.52°C

8 0
3 years ago
If your density was supposed to be 2.3 g/mL, but you calculated yours to be 2.1 g/mL, what is your percent error?
Reptile [31]

Answer: 0.08695652

Explanation:

You would do the answer you got subtracting from the expected answer over your expected answer

6 0
3 years ago
Can someone please help me with this it's for a grade.
alukav5142 [94]

Answer:

The Answer is 'D'

Explanation:

The diagram on the down side shows the behavior of the particles of a liquid so I suppose it is the ocean. While the top diagram shows the behavior of the particles of a gas so I am sure it's the air. Therefore I chose the last diagram because it describes exactly how you wanted in the question, which is the Ocean's water evaporating to become gas or the 'air' as we say

<em>Thank</em><em> </em><em>you</em><em> </em><em>and</em><em> </em><em>I</em><em> </em><em>hope</em><em> </em><em>you</em><em> </em><em>like this</em><em> </em><em>answer</em><em>! </em>

7 0
3 years ago
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