Answer:
C3H6O3
Explanation:
To find the emperical of the compound,you divide the percentage composition of each element by it's molar mass(C=12, H=1, O=16).
For C,40/12=3.33. For H,6.71/1=6.71
For O,53.19/16=3.33.
Then divide everything by the smallest value which is 3.33
For C,3.33/3.33=1. For H,6.71/3.33=~2.
For O,3.33/3.33=1.
Emperical formula=CH2O
Molecular formula=(CH2O)n=90
(12+2*1+16)n=90
30n=90, n=3
molecular formula=(CH2O)3
molecular formula of lactic acid=C3H6O3.
Halogen, any of the six nonmetallic elements that constitute Group 17 (Group VIIa) of the periodic table. The halogen elements are fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (At), and tennessine (Ts).
I hope this helps you.
Density is equal to mass divided by volume.
So D=118/12
D=9.83
17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
First, we will convert 17.8 g of NaClO₄ to moles using its molar mass (122.44 g/mol).

Next, we will convert 0.145 moles to molecules of NaClO₄ using Avogadro's number; there are 6.02 × 10²³ molecules in 1 mole of molecules.

NaClO₄ is a strong electrolyte that dissociates according to the following equation.
NaClO₄ ⇒ Na⁺ + ClO₄⁻
The molar ratio of NaClO₄ to Na⁺ is 1:1. The number of Na⁺ in 8.73 × 10²² molecules of NaClO₄ is:

The molar ratio of NaClO₄ to ClO₄⁻ is 1:1. The number of ClO₄⁻ in 8.73 × 10²² molecules of NaClO₄ is:

The molar ratio of ClO₄⁻ to Cl is 1:1. The number of Cl in 8.73 × 10²² ions of ClO₄⁻ is:

The molar ratio of ClO₄⁻ to O is 1:1. The number of O in 8.73 × 10²² ions of ClO₄⁻ is:

17.8 g of sodium perchlorate contains 8.73 × 10²² Na⁺ ions, 8.73 × 10²² ClO₄⁻ ions, 8.73 × 10²² Cl atoms and 3.49 × 10²³ O atoms.
You can learn more Avogadro's number here: brainly.com/question/13302703
Magnesium hydroxide react with hydrochloride to give magnesium chloride and water
THe word equation
magnesium hydroxide +hydrochroric acid---> Magnesium chloride+water
chemical equation
Mg(OH)2(S) + 2HCl (L)--> Mgcl2(S) + 2H2O(l)
when balancing the equation we count the number of molecules in the reactant side which should balance with molecules in the product side.