There are 2 moles of O stones present in 88 grams of CO2. Why? Well, we can find the amount of moles present in 88 grams of CO2 by dividing the mass by the molar mass. The mass of CO2 comes out to be 88 grams. The molar mass of CO2 comes out to be 44 grams. Because 88 is the mass of CO2 and 44 is the molar mass of CO2, we can divide 88 by 44 to identify that there are 2.0 moles of O atoms present in 88 grams of CO2.
Your final answer: There are 2.0 moles of O atoms present in 88 grams of CO2. Your final answer to this question is D, or 2.0 moles. If you need to better understand, let me know and I will gladly assist you.
Answer:
Total 5 significant digits.
Explanation:
Significant digits are the numbers that give a meaningful contribution. For example, digit 013 has the 2 significant digits and zero is not a significant digit because digits 1 and 3 give meaningful contribution but digit zero does not value meaningful contribution. Similarly, the 89015 has a total of 5 significant digits and these digits are the 8, 9, 0, 1, and 5.
The molality is calculated using the following rule:
molality = number of moles of solute / kg of solvent
From the periodic table:
molar mass of lithium = 6.941 gm
molar mass of chlorine = 35.453 gm
molar mass of LiCl = 6.941 + 35.453 = 42.394 gm
number of moles found in 42 gm = mass / molar mass = 42 / 42.394 = 0.99
molality = 0.99 / 3.6 = 0.275 m
Magnet to remove the iron, filter paper (or some filter that could trap sand) to remove the sand, and heat up the salt and water, causing the water to evaporate to separate them.
