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3 years ago

What is the percent error when a student measures the volume to be 19.3 liters when

Chemistry

1 answer:

katrin2010 [14]3 years ago

7 0

Answer:

The answer is

Explanation:

The percentage error of a certain measurement can be found by using the formula

$P(\%) = \frac{error}{actual \: \: number} \times 100\% \\$

From the question

actual volume = 22.4 L

error = 22.4 - 19.3 = 3.1

The percentage error is

$P(\%) = \frac{3.1}{22.4} \times 100 \\ = 13.839285714...$

We have the final answer as

Hope this helps you

You might be interested in

Calculate the ph of sweater which has a hydrogen ion concentration of 1 times 10 to the power of 8

telo118 [61]

Answer:

The pH of the sweater containing Hydrogen ion concentration

$[H^{+}]=1\times 10^{-8}$ is

<u></u>

Explanation:

pH = It is the negative logarithm of activity (concentration) of hydrogen ions.

pH = -log([H+])

Now, In the question the concentration of [H+] ions is :

$[H^{+}]=1\times 10^{-8}$

$pH = -log(1\times 10^{-8})$

use the relation:

$log(10^{-a})=a$

$pH= -(-8)$

pH = 8

Note : <em><u> 1 times 10 to the power of 8 must be" 1 times 10 to the power of -8"</u></em>

If the concentration is

$[H^{+}]=1\times 10^{8}$

Then pH = -8 , which is not possible . So in that case the pH calculation is by other method

5 0

3 years ago

You are required to prepare 500 ml of a 6.00 M solution of HNO3 from a stock solution of 12.0 M. Describe in detail how you woul

andriy [413]

Answer: 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.

Explanation:

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = concentration of stock solution = 12.0 M

$V_1$ = volume of stock solution = ?

$C_2$ = concentration of diluted solution= 6.00 M

$V_2$ = volume of diluted acid solution = 500 ml

Putting in the values we get:

$12.0\times V_1=6.00\times 500$

$V_1=250ml$

Thus 250 ml of stock solution with molarity of 12.0 M is measured using a pipette and 250 ml of water is added to volumetric flask of 500 ml to make the final volume of 500 ml.

8 0

3 years ago

What are two ways that substances can pass through a cell membrane out of the cell

Paha777 [63]

Answer:

Passive transport, and Active transport

Explanation:

Do u need an explanation?

If so you can tell me and I'll edit it

Hope this helped a little!

3 0

3 years ago

Compound A melts at 801o C and is soluble

Karo-lina-s [1.5K]

Explanation:

Organic compounds are defined as the compounds which contain carbon as their main element. For example, $CH_{3}OH$ is an organic compound.

Generally, organic compounds are non-polar in nature and due to the presence of covalent bonding organic compounds have low melting point.

As compound A melts at $801^{o}C$ and is soluble in water. This means it is an ionic compound as it has high melting point and it is also polar in nature.

Whereas compound B melts at $24^{o}C$ and is insoluble in water. This means that this compound has covalent bonding and it is also non-polar in nature . Hence, it is more likely to be organic in nature.

Thus, we can conclude that compound B is more likely to be an organic

compound.

7 0

3 years ago

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

meriva

Answer:

$\large \boxed{\text{77.4 mL}}$

Explanation:

Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

V/mL: 249

c/mol·L⁻¹: 0.0443 0.285

1. Calculate the moles of Ba(OH)₂

$\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}$

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

$\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1 mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}$

3. Calculate the volume of HCl

$V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}$

8 0

3 years ago