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Brilliant_brown [7]
3 years ago
13

A 3.42 gram sample of an unknown gas is found to occupy a volume of 1.90 L at a pressure of 547 mm Hg and a temperature of 33 °C

. The molecular weight of the unknown gas is ___ g/mol.
Chemistry
1 answer:
liq [111]3 years ago
3 0
Use the Ideal Gas Law to find the moles of gas first.

Be sure to convert T from Celsius to Kelvin by adding 273.

Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.

PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K

n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas

Now divide grams by mol to get the molecular weight.

3.42 g / 0.0545 mol = 62.8 g/mol
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5 0
2 years ago
A sample of a vapor occupies a volume of 500 mL at 65°C. If pressure remains constant, what is the volume of the gas at standard
DedPeter [7]

Answer:

403 mL

Explanation:

First, I will assume that the mole is 1, because you are not specifing this.

Now, with the innitial data, we need to get the pressure:

T = 65+273 = 338 K

V = 500 / 1000 = 0.5 L

Now if:

PV = nRT

Then:

P = nRT/V   and V = nRT/P

Let's calculate the P:

P = 1 * 0.082 * 338 / 0.5 = 55.432 atm

The standard temperature is 0° C or 273 K so, the volume is:

V = 1 * 0.082 * 273 / 55.432

V = 0.40384 L or simply 403.84 mL

8 0
3 years ago
1. Agriculture most likely began
777dan777 [17]

Answer:

Option D - 10,000 years ago in the Middle East

4 0
3 years ago
Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to
vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

   H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)

Next we will calculate how many moles of H_2SO_4 are present in 85.00 mL of 1.500 M sulfuric acid.

As,       Molarity = \frac{\text{moles of solute}}{\text{liters of solution&#10;}}

            1.500 M = \frac{n}{0.08500 L&#10;}

                    n = 0.1275 mol H_2SO_4

Now set up and solve a stoichiometric conversion from moles of H_2SO_4  to grams of NaHCO_3. As, the molar mass of NaHCO_3 is 84.01 g/mol.

 0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})

                 = 21.42 g NaHCO_3

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0
3 years ago
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