Answer:
Mass of KNO3= 10g
Molar mass of KNO3 = 101.1032g/mol
Volume = 250ml = 0.25L
No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3
no of mole of KNO3 = 10/101.1032
No of mole of KNO3 = 0.09891
molarity of KNO3 = no of mole of KNO3/Vol (L)
Molarity = 0.09891/0.25 = 0.3956M
Molarity of KNO3 = 0.3956M
Percentage yield = actual yield/theoretical yield
So divide 21.0 g by 22.7 g and multiply it by 100 to find the percentage yield
Thank you for posting your math problem here. To convert 3.9x10^5mg to dg the answer is <span>3.9 x 10^3 dg. Below is the solution:
Solution:
</span><span>1mg=0.01dg
</span><span> dg= 3.9 X 10^5mg
</span>dg = <span>(3.9 X 10^5) x 0.01
dg = </span><span>3.9 x 10^3 </span>
Answer:
260.34g
Explanation:
First, you need to know what angelic acid is comprised of. It is written as C₅H₈O₂.
In order to solve for the mass of 2.6 moles of angelic acid, you need the mass of 1 mole of angelic acid. This can be found by adding the masses from the periodic table, like shown below:
5 carbon atoms = (5)(12.01g) = 60.05g
8 hydrogen atoms = (8)(1.01) = 8.08g
2 oxygen atoms = (2)(16) = 32g
angelic acid = 60.05 + 8.08 + 32 = 100.13g
Then, set up a basic stoichiometric equation and solve. The units should cancel out.

We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:
osmotic pressure = CRT
<span>C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L</span>
<span>moles of particles = C*V = 0.3189*0.250 =0.0797 mol </span>
<span>0.0797 = moles of sucrose + 2*moles of salt </span>
<span>x + 2y = 0.0797 </span>
<span>and </span>
<span>x(MMsucrose) + y(MMNaCl) = 10.2</span>
<span>342x + 58.5y = 10.2
</span>
<span>solve for x and y
</span>
<span>x = 0.0252 mol sucrose</span>
<span>y = 0.0273 mol NaCl
</span>
<span>mass Sucrose = 0.0252(342) = 8.6184 g </span>
<span>mass NaCl = 0.0273(58.5) = 1.5971 g </span>
<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>