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:
4.62
so 5
the ratio is 2 na chlorates for 3 O2 so multiply 7 by 2/3
Explanation:
Answer:
c. the heart
all the rest actually will excrete, the heart is in a closed system so contributes to excretory organs but doesn't actually do it barring an accident
Explanation:
Answer:
16.8 g of AgCl are produced
Explanation:
The reactants are: NaCl and AgNO₃
The products are: AgCl, NaNO₃
Balanced equation: NaCl(aq) + AgNO₃(aq) → NaNO₃(aq) + AgCl(s) ↓
We convert the mass of AgNO₃ to moles → 10 g / 85g/mol = 0.117 moles
Ratio is 1:1, therefore 0.117 moles of nitrate will produce 0.117 moles of AgCl.
According to stoichiormetry.
We convert the moles to mass → 0.117 mol . 143.3g /1mol = 16.8 g
