The elements that fill the d orbitals on the periodic table are called transition metals.
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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Answer:
option b. B3+
Explanation:
Boron takes the 5th position on the periodic table, therefore it has 5 electrons....2 on the inside and 3 on the outside. when it lost it 3 external electrons, it become positively charged with the amount of electron it loses.
Answer:
Volume of the solutions
This is the most important factor for her to control.
39.25 g of water (H₂O)
Explanation:
We have the following chemical reaction:
2 H₂ + O₂ → 2 H₂O
Now we calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of H₂ = 14.8 / 2 = 7.4 moles
number of moles of O₂ = 34.8 / 32 = 1.09 moles
We see from the chemical reaction that 2 moles of H₂ will react with 1 mole of O₂ so 7.4 moles of H₂ will react with 3.7 moles of O₂ but we only have 1.09 moles of O₂ available. The O₂ will be the limiting reactant. Knowing this we devise the following reasoning:
if 1 moles of O₂ produces 2 moles of H₂O
then 1.09 moles of O₂ produces X moles of H₂O
X = (1.09 × 2) / 1 = 2.18 moles of H₂O
mass = number of moles × molar weight
mass of H₂O = 2.18 × 18 = 39.25 g
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limiting reactant
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