2. How much potassium chlorate is needed to produce 20.0 mL of oxygen at 670 mm Hg and 20 degrees C.

3. A certain compound containing only carbon and hydrogen was found to have a vapor density of 2.550 g/L at 100 degrees C and 760 mm Hg. If the empirical formula of this compound is CH, what is the molecular formula of this compound?

To answer these questions, we need to use the ideal gas law and the concept of molar mass.

2. To determine the amount of potassium chlorate needed, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, we need to convert the given volume from mL to L:
20.0 mL = 20.0 mL * (1 L / 1000 mL) = 0.020 L

Next, we need to convert the given pressure from mm Hg to atm:
670 mm Hg = 670 mm Hg * (1 atm / 760 mm Hg) = 0.8829 atm

Now, let's substitute the given values into the equation:
(0.8829 atm) * (0.020 L) = n * (0.0821 L * atm / K * mol) * (20 + 273 K)

Simplifying and solving for n:
0.017658 = n * (0.0821 * 293)
n = 0.017658 / (0.0821 * 293)
n ≈ 7.1885 x 10^-4 mol

Since the balanced equation for the decomposition of potassium chlorate is:
2KClO3(s) → 2KCl(s) + 3O2(g)
We can see that for every 2 moles of KClO3, we get 3 moles of O2.

Therefore, to produce 7.1885 x 10^-4 mol of O2, we need:
(7.1885 x 10^-4 mol O2) * (2 mol KClO3 / 3 mol O2) = 4.7923 x 10^-4 mol KClO3

To find the mass of KClO3 needed, we need to multiply the number of moles by its molar mass. The molar mass of KClO3 is calculated as follows:
(Molar mass of K) + (Molar mass of Cl) + 3*(Molar mass of O)
(39.10 g/mol) + (35.45 g/mol) + 3*(16.00 g/mol) = 122.55 g/mol

Therefore, the mass of KClO3 needed is:
(4.7923 x 10^-4 mol KClO3) * (122.55 g/mol) = 0.0587 g KClO3

So, approximately 0.0587 grams of potassium chlorate is needed to produce 20.0 mL of oxygen at 670 mm Hg and 20 degrees C.

3. To find the molecular formula of the compound, we need to use the concept of molar mass and compare it to the empirical formula.

The empirical formula of CH tells us that for every one mole of compound, we have one mole of carbon (C) and one mole of hydrogen (H).

The molar mass of CH is calculated as follows:
(Molar mass of C) + (Molar mass of H)
(12.01 g/mol) + (1.01 g/mol) = 13.02 g/mol

The given vapor density of 2.550 g/L at 100 degrees C and 760 mm Hg can be converted to molar mass as follows:
2.550 g/L = 2.550 g/L / (22.4 L/mol) = 0.114 mol/L

Now, we need to calculate the molar mass of the compound by multiplying the empirical formula molar mass by a whole number (n):
n * 13.02 g/mol = 0.114 mol/L

Solving for n:
n = 0.114 mol/L / 13.02 g/mol
n ≈ 0.0088

Since n is approximately 0.0088, it tells us that the molecular formula must be a multiple of the empirical formula. To find the molecular formula, we can divide the molar mass of the compound by the molar mass of the empirical formula:

Molar mass of the compound / Molar mass of CH = 13.02 g/mol / 13.02 g/mol = 1

Therefore, the molecular formula of the compound is CH.

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