1.75 mole CHCl3 (6.02 X 10^23 mole CHCl3)
= 1.05 X 1024 atoms of CHCl3
====================
You must appy the principle of mass to the speed and tempo of gravity. Then you should get the molar mass of the elements at hand and then divide it by 2. Your answer should be .88g.
1.54g
175 g Ca (1 mol / 40.078 g) (6.022x10^23 atoms / 1 mol) = 2.63x10^24 Ca atoms There are 2.63x10^24 c.alcium atoms in 175 grams of calcium.
The mass of one gold atom is approximately 197 atomic mass units. Therefore, the mass of 175 gold atoms would be approximately 34,475 atomic mass units.
To find the molar mass of CHCl3, you would calculate the sum of the atomic masses of carbon, hydrogen, and three chlorine atoms. The atomic masses are: carbon (C) = 12.01 g/mol, hydrogen (H) = 1.01 g/mol, and chlorine (Cl) = 35.45 g/mol. Therefore, the molar mass of CHCl3 is 12.01 (C) + 1.01 (H) + 3*(35.45) (3 Cl) = 119.37 g/mol.
A mol = 6,022x10^23 atoms 1,5 mol = 9,033x10^23 atoms Always.
The molar mass of chloroform (CHCl3) is approximately 119.38 g/mol.
175 g Ca (1 mol / 40.078 g) (6.022x10^23 atoms / 1 mol) = 2.63x10^24 Ca atoms There are 2.63x10^24 c.alcium atoms in 175 grams of calcium.
The mass of one gold atom is approximately 197 atomic mass units. Therefore, the mass of 175 gold atoms would be approximately 34,475 atomic mass units.
the constant Mole (mol): 6.02 x 10^23 are how many atoms you have per mol so the answer can be 7 mol atoms or 6.02 x 10^23 atoms per mol x 7 actual answer is 4.214 X10^24 atoms in 7 mol
A mol = 6,022x10^23 atoms 1,5 mol = 9,033x10^23 atoms Always.
There are approximately 1.21 x 10^23 lead (Pb) atoms in 0.200 mol of Pb. This number is calculated using Avogadro's number, which is 6.022 x 10^23 atoms/mol.
0.45 mol BaSO4 have 2,698.10e23 sulfur atoms.
There are 1.41 e24 atoms of Na in 2.35 mol of NaCl.
1.3 mol
481,7713432.1023 atoms
The answer is 18,061.1023 atoms.
48,177 134 32.1023 atoms
Avagadro's number = 6.022 × 1023 atoms/mol0.5 mol × (6.022 × 1023) atoms/mol = 3.011 ×1023 atoms