How many grams of barium sulfate form when 35.0 ml of 0.160 m barium chloride reacts with 58.0 ml of 0.065 m sodium sulfate? do
1 answer:
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Answer:
no. of water molecules associated to each molecule of = 4
Explanation:
Mass of before heating = 19.8 g
Mass of after heating = 12.6 g
Difference in mass of before and after heating
= 19.8 - 12.6 = 7.2 g
Difference in mass corresponds to mass of water driven out.
Molar mass of water = 18 g/mol
No. of moles of water =
Mass of obtained after heating is mass of anhydrous
.
Mass of anhydrous = 12.6 g
Molar mass of = 125.9 g/mol
No. of mol of anhydrous =
so,
0.1 mol of have 0.4 mol of water
1 mol of will have =
Hence, no. of water molecules associated to each molecule of = 4
This is a combined gas law problem, according to which
where P is the pressure of the gas, V is the volume of the gas, and T is the temperature of the gas, and the subscripts 1 and 2 correspond to the initial and final conditions of the gas. In this problem, we are given the initial pressure, volume, and temperature of the gas in the balloon:
P₁ = 1.0 atm
V₁ = 1.8 L
T₁ = 295.15 K (K = °C + 273.15).
Moreover, we are given the final pressure and temperature of the gas in the balloon.
P₂ = 0.86 atm
T₂ = 281.15 K.
What we want to find is the final volume, V₂, which we can obtain by rearranging the combined gas equation to solve for V₂:
This answer has three significant figures. However, the question as written would warrant an answer that comprises one significant figure (as 8 °C has only one sig fig). In that case, the answer would be 2 L. If the answer is to be given to two significant figures, the volume would then be 2.0 L.