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>
Answer:
Inspiration
Explanation:
This question is on application of Boyle's law; <u>pressure is inversely proportional to volume</u>.when we inhale air, the diaphragm and the muscles in the ribs contract thus increasing the volume in the lungs.Increased volume of the lungs cause the pressure to decrease.During exhaling, the diaphragm and muscles in the ribs relax, making the lungs to recoil and reduce in volume to force air out.Pressure in the lungs is increased than that in the environment making air to move out.
A. The radioactive decay equation is N = N0
where T is the
half-life (5730 years), N0 is the number of atoms at time t = 0 and
N is the number at time t.
Rewriting this as:
(N/N0) =
Since N = (1/8) N0 and
substituting known values:
1/8 = 
Taking ln of both
sides:
ln(1/8)= -ln(2)*t/5730
t = - 5730 * ln(1/8) /
ln (2)
t = 17,190 years
The tree was cut down 17,190
years ago.
B. N0 = 1,500,000 carbon-14 atoms
Since N = (1/8) N0
N = 187,500 carbon
atoms left
Answer:
we have two loops in our body in which blood circulates. One is oxygenated, meaning oxygen rich, and the other is deoxygenated, which means it has little to no oxygen, but a lot of carbon dioxide.
