The problem you have written you almost have it solved. Take the moles that you have calculated and multiply that by the molecular weight to get the grams.
The STP problem:
use the moles you calculated along with 1 atm for Pressure, and 273 for the temperature and plug into the PV = nRT equation. (also use 0.0821 for R)
From there you can solve for the volume
Hope this helps!
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Spectrophotometric cell or a cuvette is made of quartz for UV spectrophotometers. These cuvettes are used as sample holders for the spectrophotometric determination of the analytes. The material that makes up the cuvette and the condition of the cuvette is to be taken care of in order to avoid erroneous absorbance readings. The sample holder or the cuvette must be removed from the spectrophotometer in between two successive readings. This is to ensure that the light sensing detector of the instrument is not affected.
Answer:
53.85%
Explanation:
Data obtained from the question include:
Mass of antimony (Sb) = 27.6g
Mass of Fluorine (F) = 32.2g
Mass of compound = 59.8g
Percentage composition of fluorine (F) =..?
The percentage composition of fluorine can be obtained as follow:
Percentage composition of fluorine = mass of fluorine/mass of compound x 100
Percentage composition of fluorine = 32.2/59.8 x 100
= 53.85%
Therefore, the percentage composition of fluorine in the compound is 53.85%
Molality can be expressed by moles of solute over
kilograms of solvent. The question asks the molality of 0.25m NaCl. 0.25m NaCl
is equal to 0.25 moles of NaCl over 1 kg of water.
