The correct answer is Cl.
Chlorine is a substance that's employed in industry and is present in a number of household goods. There are times when chlorine takes the form of toxic gas. Chlorine gas can be converted into a liquid by applying pressure and cooling so that it can be transported and stored. The term "oxidation number" refers to the number of electrons that an element's atom either loses or gains during the production of a compound. The charge that an atom seems to have when forming ionic connections with other heteroatoms is used to define an atom's oxidation number. Even if it develops a covalent bond, an atom with a higher electronegativity is given a negative oxidation state.
Learn more about oxidation numbers here:-
brainly.com/question/10079361
#SPJ4
a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>
Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

volume NO at 1273 K and 1 atm

b. 15 L NH3 at STP ( 1mol = 22.4 L)

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

mass H2O(MW = 18 g/mol) :

c. mol NO at 1273 K and 1 atm :

mol ratio of NO : O2 = 4 : 5, so mol O2 :

Volume O2 at STP :

Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.
Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>