All categories

3 years ago

If you find the mass of a sample to be 42.1 g and the actual mass is 42.0 g. What wold be the percent error

Chemistry

1 answer:

Ksivusya [100]3 years ago

4 0

Answer:

0.24%

Explanation:

The following data were obtained from the question:

Measured value = 42.1 g

Actual value = 42 g

Percentage error =?

Percentage error can be obtained by using the following formula:

Percentage error = |Measured value – Actual value | / Actual value × 100

With the above formula, we can obtain the percentage error as follow:

Measured value = 42.1 g

Actual value = 42 g

Percentage error =?

Percentage error = |42.1 – 42| / 42 × 100

Percentage error = 0.1/42 × 100

Percentage error = 10/42

Percentage error = 0.24%

Therefore, the percentage error is 0.24%

You might be interested in

1. What is erosion? Provide at least one example of a landform made by erosion.

zepelin [54]

I believe that the subject is supposed to be science

5 0

3 years ago

Read 2 more answers

Explain why are all valence electrons are either in an s or p orbital.

ahrayia [7]

They are two representatives of the sublevels/energy levels.

4 0

3 years ago

An atom has a mass number of 41 and contains 21 neutrons. which element is it?

Kipish [7]

The element you are looking for is Scandium, the 4th element in the third group.

6 0

3 years ago

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold

Sonja [21]

Answer:

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

Explanation:

The formula we are going to use is:

$N_{Au}=\frac{N_A*C_{Au}}{{\frac{C_{Au}A_{Au} }{\rho_{Au}}}+\frac{A_{Au}}{\rho_{Ag}}(100-C_{Au})}$

Where:

$N_{Au}$ are number of gold atoms.

$N_A$ is Avogadro Number.

$C_{Au}$ is the amount of gold.

$A_{Au}$ is the atomic weight of gold.

$\rho_{Au}$ is the density of gold.

$\rho_{Ag}$ is the density of silver.

$C_{Ag}$ is the amount of silver.

$N_{Au}=\frac{6.023*10^{23}*45\%wt}{\frac{45\%wt*196.97}{19.32}+\frac{196.97}{10.49}(100-45\%wt)}\\ N_{Au}=1.81718*10^{22}\ atoms/cm^3$

The Number of gold atoms are = $N_{Au}=1.81718*10^{22}\ atoms/cm^3$

5 0

3 years ago

How many moles of each element are in one mole of Be(OH)2?

blondinia [14]

Answer:

Explanation:

To find the number of moles of each element present in 1 mole of Be(OH)₂, we use the ratio of the molar masses of the elements to the compound and multiply by the given mole.

Molar mass of Be(OH)₂ = 9 + 2(16+1) = 43g/mol

number of moles of Be = $\frac{9}{43}$ x 1 = 0.209mole

number of moles of O = $\frac{32}{43}$ x 1 = 0.744mole

number of moles of H = $\frac{2}{43}$ x 1 = 0.047mole

3 0

3 years ago