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
73.606 °C is the freezing point of the solution made with with 1.31 mol of CHCl3 in 530.0 g of CCl4.
Explanation:
Data given:
number of moles of CHCl3 = 1.31 moles
mass of solvent CHCl3 = 530 grams or 0.53 kg
Kf = 29.8 degrees C/m
freezing point of pure solvent or CCl4 = -22.9 degrees
freezing point = ?
The formula used to calculate the freezing point of the mixture is
ΔT = iKf.m
m= molality
molality = 
putting the value in the equation:
molality= 
= 2.47 M
Putting the values in freezing point equation
ΔT = 1.31 x 29.8 x 2.47
ΔT = 73.606 degrees
No,they just separate away from each other
<h2>Answer:</h2>
2 hydrogen atoms.
<h2>Explanations:</h2>
Given the chemical formula H2SO4,
The compound shows that the formula has 4 atoms of oxygen, one atom of sulfur and 2 hydrogen atoms.
Therefore, the number of hydrogen atom in the molecule H2SO4 is 2 hydrogen atoms.
Answer:
the first one is the answer