The correct option is (C) 6.02 X 10²³
A sample of CH₄O with a mass of 32.0 g contains <u>6.02 X 10²³</u> molecules of CH₄O.
To calculate the number of moles;
Molar mass of CH₄O = C + 4(H) + O
= 12.01 + 4(1.008) + 16
= 32.04 g/mol
So, 1 mol of CH₄O = 32.04 g of CH₄O
Given, 32.0 g of CH₄O
According to Avagadro's constant 1 mole of a substance contains 6.022× 10^23 particles (molecules, atoms or ions).
= (32.0 g/1)(1 mol CH₄O/32.04 g CH₄O)(6.02x10²³/1 mol CH₄O)
= 6.02 X 10²³ molecules of CH₄O
Hence, a sample of will contain number of molecules 6.02 X 10²³ molecules.
Learn more about the Moles calculation with the help of the given link:
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A
Explanation: hydrogen has a charge of 1 and oxygen has a charge of -2, so you will need 2 hydrogens to balance out the oxygen
Answer:
Scandium
Explanation:
Mendeleev played an important role in the development of the modern periodic table. His periodic table was filled with gaps. He said that these gaps were elements that were yet to be discovered. He rightly predicted many elements which have now been discovered and fitted in their proper places in the periodic table.
He used the prefix ''eka'' to refer to elements whose properties were alike but were yet to be discovered at that time.
The compound named ekaboron which he predicted to have an atomic weight between 65 (zinc) and 75 (arsenic) with a valence similar to aluminum was later discovered in 1879 and properly named scandium.
Answer:
The IHD = 0
There is no unsaturation in the compound C4H9Cl
Explanation:
IHD = Index for Hydrogen Deficiency .It determines total number of unsaturation or Pi- bond present in the molecule.It is calculated using the formula:
For Molecule:

Here . C = carbon
H = Hydrogen
O = Oxygen
N = Nitrogen
X = Halogen
The IHD is calculated as :
...............(1)
For example :
C4H8O2
Here c = 4 , h = 8 , o = 2 .
n = x=0
Put the value in equation (1) and find IHD



IHD = 1
For C4H9Cl
c = 4 , h = 9 and x = 1
IHD for this molecule can be calculated as :



= 0
This means there is no unsaturation in the molecule.
Each bond is sigma bond and properly saturated.