<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
A. Phase changing. When phase changes nothing chemically changes about the substance, its still the same thing.
Answer : Option D) The atmospheric conditions vary as one changes latitude and altitude.
Explanation : The composition of the atmosphere varies according to the latitude and altitude because of the unequal heating of the earth surface at different latitudes and altitudes which results into atmospheric changes. It also creates different regions and zones.
<span>Based on the experience of the responder, to correctly calculate measurements in real-world. Firstly is to avoid errors as much as possible. Errors are what makes your measurement invalid and unreliable. There are two types of error which is called the systematic error and the random error. Each error has different sources. Words that were mentioned –invalid and unreliable are very important key aspects to determine that your measure is truly accurate and consistent. Some would recommend using the mean method, doing three trials in measuring and getting their mean, in response to this problem.</span>
Atomic Number of Zinc is 30, means it contains 30 electrons. So, its electronic configuration is as follow,
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰
As,
1s², 2s², 2p⁶, 3s², 3p⁶ = Argon
So,
Electronic configuration of Zinc in shorthand notation is as follow,
[Ar] 4s², 3d¹⁰
