<span>Step 1 is to determine the mass of each part
Mass of Ca is 40.08 g
Mass of C is 12.01 g
Mass of O is 16.00 x 3 = 48.00 g
Step 2 is to determine the total mass of the compound
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g
Step 3 is to determine the % of each part using the following formula:
Mass of part / total mass x 100 =
40.08 / 100.09 x 100 = 40.04 % Ca
12.01 / 100.09 x 100 = 12.00 % C
48.00 / 100.09 x 100 = 47.96 % O
Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :)
40.04
+12.00
+47.96
=100.00</span><span>
</span>
Answer:
THE VOLUME OF THE NITROGEN GAS AT 2.5 MOLES , 1.75 ATM AND 475 K IS 55.64 L
Explanation:
Using the ideal gas equation
PV = nRT
P = 1.75 atm
n = 2.5 moles
T = 475 K
R = 0.082 L atm/mol K
V = unknown
Substituting the variables into the equation we have:
V = nRT / P
V = 2.5 * 0.082 * 475 / 1.75
V = 97.375 / 1.75
V = 55.64 L
The volume of the 2.5 moles of nitrogen gas exerted by 1.75 atm at 475 K is 55.64 L
Three classes: 28×3=84 students
8 test tubes per student: 84× 8= 672
hope this helps!
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)
