They are called representative elements.
The main constituent of gallstones is cholesterol. Cholesterol may have a role in heart attacks and blood clot formation. Its elemental percentage composition is 83.87% C, 11.99% H, and 4.14% O. It has a molecular weight of 386.64 amu. Empirical formula is C₃H₄O₁ and Molecular formula is 7(C₃H₄O₁).
<h3>What is Empirical Formula ?</h3>
Empirical formula is the simplest whole number ratio of atoms present in given compound.
Element % Atomic mass Relative no. of atoms Simplest whole ratio
C 83.87 12
= 6.98
= 3
H 11.99 1
= 11.09
= 4
O 4.14 16
= 0.25
= 1
Thus the empirical formula is C₃H₄O₁.
<h3>How to find the Molecular formula of compound ?</h3>
Molecular formula = Empirical formula × n
n = 
= 
= 7
Molecular formula = Empirical formula × n
= 7 (C₃H₄O₁)
Thus from the above conclusion we can say that The main constituent of gallstones is cholesterol. Cholesterol may have a role in heart attacks and blood clot formation. Its elemental percentage composition is 83.87% C, 11.99% H, and 4.14% O. It has a molecular weight of 386.64 amu. Empirical formula is C₃H₄O₁ and Molecular formula is 7(C₃H₄O₁).
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The pH meter measure the concentration of the hydrogen(+) ion, and the concentration of Hydroxide(-). Hydrogen is acidic, Hydroxide is basic.
The answer is D.
<span>1.40 x 10^5 kilograms of calcium oxide
The reaction looks like
SO2 + CaO => CaSO3
First, determine the mass of sulfur in the coal
5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4
Now lookup the atomic weights of Sulfur, Calcium, and Oxygen.
Sulfur = 32.065
Calcium = 40.078
Oxygen = 15.999
Calculate the molar mass of CaO
CaO = 40.078 + 15.999 = 56.077
Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight.
8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles
Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass
2.49 x 10^3 * 56.077 = 1.40 x 10^5
So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
The answer is , C. Both the atomic mass and the atomic number increase from left to right .