A. NaCl(s) and O2(g)
B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)
C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3
D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)
E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl
F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2
G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2
H. Percent yield = 10/45.1 • 100% = 22.2% yield
Answer: A volume of 500 mL water is required to prepare 0.1 M 
from 100 ml of 0.5 M solution.
Explanation:
Given: 
= 0.1 M, 
= ?
= 0.5 M, 
= 100 mL
Formula used to calculate the volume of water is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that a volume of 500 mL water is required to prepare 0.1 M from 100 ml of 0.5 M solution.
Formula for hydroselenic acid: H2Se
Answer:
B: Carbon Dioxide emissions from fossil fuels
The answer is c. 104 g because 152-64 equals 88 and 192-88 equals 105
