<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
Answer: B) The trees where peppered moths rest become lighter in color.
The mean is simply the arithmetic average of all your raw data. This can be solved methodically by summing up all of the raw data points that you have. Take note how many raw data points you used, because this will be used to divide the sum. You will obtain the mean.
Answer:
25.11 g.
Explanation:
- It is clear from the balanced equation:
<em>Ag₂O + 2HCl → 2AgCl + H₂O.</em>
<em></em>
that 1.0 mole of Ag₂O reacts with 2.0 moles of HCl to produce 2.0 mole of AgCl and 1.0 moles of H₂O.
- 7.8 g of HCl reacts with excess Ag₂O. To calculate the no. of grams of Ag₂O that reacted, we should calculate the no. of moles of HCl:
<em>no. of moles of HCl = mass/atomic mass</em> = (7.9 g)/(36.46 g/mol) = <em>0.2167 mol.</em>
- From the balanced equation; every 1.0 mol of Ag₂O reacts with 2 moles of HCl.
∴ 0.2167 mol of HCl will react with (0.2617 mol / 2 = 0.1083 mol) of Ag₂O.
<em>∴ The mass of reacted Ag₂O = no. of moles x molar mas</em>s = (0.1083 mol)(231.735 g/mol) = <em>25.11 g.</em>
For this problem, we should use the Henry's Law formula which is written below:
P = kC
where
P is the partial pressure of the gas
k is the Henry's Law constant at a certain temperature
C is the concentration
Substituting the values,
1.71 atm = (7.9×10⁻⁴<span> /atm)C
Solving for C,
C = 2164.56 molal or 2164.56 mol/kgwater
Let's make use of density of water (</span>1 kg/1 m³) and the molar mass of NF₃ (71 g/mol).<span>
Mass of NF</span>₃ = 2164.56 mol/kg water * 1 kg/1 m³ * 1 m³/1000000 mL * 150 mL * 71 g/mol = 23.05 g
