When utilizing the gravimetric method, it is crucial to completely dissolve your sample in 10 mL of water. A quantitative technique called gravimetric analysis employs the selective precipitation of the component under study from an aqueous solution.
A group of techniques known as gravimetric analysis are employed in analytical chemistry to quantify an analyte based on its mass. Gravimetric analysis is a quantitative chemical analysis technique that transforms the desired ingredient into a substance (of known composition) that can be extracted from the sample and weighed. This is a crucial point to remember.
Gravimetric water content (g) is therefore defined as the mass of water per mass of dry soil. To calculate it, weigh a sample of wet soil, dry it to remove the water, and then weigh the dried soil (mdry). Dimensions of the sample Water is commonly forgotten despite having a density close to one.
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Question:
At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?
Answer:
The new pressure is: 1.896 atm
Explanation:
At standard temperature and pressure, we have:
Outside, we have:
Required
Determine the new pressure
Using combined gas law, we have:
This gives:
Solve for 
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
Answer:
Mass percentage → 0.074 %
[F⁻] = 741 ppm
Explanation:
Aqueous solution of flouride → [F⁻] = 0.0390 M
It means that in 1L of solution, we have 0.0390 moles of F⁻
We need the mass of solution and the mass of 0.0390 moles of F⁻
Mass of solution can be determined by density:
1g/mL = Mass of solution / 1000 mL
Note: 1L = 1000mL
Mass of solution: 1000 g
Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g
Mass percentage → (Mass of solute / Mass of solution) . 100
(0.741 g / 1000 g) . 100 = 0.074 %
Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)
0.741 g . 1000 mg/1g = 741 mg
1000 g . 1 kg/1000 g = 1kg
741 mg/1kg = 741 ppm
