Answer:
44.8 litres is the answer to the question
Answer:
24.7 amu
Explanation:
An isotope is when an element can have different number of neutrons but they have same number of protons.
In order to calculate the average atomic mass with the given information do the following operations:
First change de percentages to fractional numbers, divide by 100.
I like to make a table, to organize all data and I believe is easier to understand.
65/100 = 0.65
35/100 = 0.35
% fraction
65.0 0.65
35.0 0.35
total100.0 1
Now multiply each mass with their corresponding fraction
24 (0.65) = 15.6
26 (0.35) = 9.1
% fraction uma uma
65.0 0.65 24 15.6
35.0 0.35 26 9.1
total100.0 1 24.7
Finally you add the resulting mass and the units will be in uma.
15.6+9.1 = 24.7
Therefore the average atomic mass of this element will be 24.7 uma.
Check the table in the document attached
C. because it doesn't break down without chemicals, so it would eventually just pile up.
Answer:
- <em>The average mass of calcium in each sample is: </em><u>0.978 g</u>
<em />
- <em>The absolute uncertainty is: </em><u>0.008 g</u>
Explanation:
The <em>absolute uncertainty </em>of the total samples indicated in the statement is ± 0.1 g.
When you multiply or divide quantities with uncertainties, you calculate the final uncertanty by adding the <em>relative uncertainties</em> together.
The relative uncertainty is the absolute uncertainty divided by the quantity:
- Relative uncertainty = 0.1g / 12.2 g = 0.008
The average mass of calcium is calculated using proportions, along with the molar masses:
- Molar mass of calcium: 40.078 g/ mol (from a periodic table)
- Molar mass of calcite: 100.085 g/mol (given)
Proportion:
- 40.078 g of calcium / 100.085 g of calcite = x / 12.2 g of calcite
- x = 12.2 × 40.078 / 100.085 g = 4.89 g calcium
So the total mass of calcium in the five samples is 4.89 g, and the average mass in each sample is:
- Average mass = total mass of five samples / number of samples
- Average mass = 4.89 g / 5 = <u>0.978 g of calcium</u>
So, the first answer is that the average mass of calcium in each sample is 0.978 g ( keep 3 signficant figures, such as the quntitiy 12.2 shows, as you have only used multiplication and division).
The absolute uncertainty of each sample is the relative uncertainty multiplied by the average mass of calcium of the five samples, rounded to one decimal:
- Absolute uncertainty = 0.978 g × 0.008 ≈ 0.008 g
The answer to the secon question is that the absolute uncertaingy of calcium in each sample is 0.008 g.