Answer:
B C A
Explanation:
because as you pour the blue stuff in it will straight up go to the bottom (and has less blue coloring) Then it will go to the top (lil more blue coloring) then eventually the whole jar will be blue
hope this helps you out bro
Physical Properties<span>: </span>Physical properties<span> can be observed or measured without changing the composition of matter. </span>Physical properties<span> are used to observe and describe matter. so physical changes are the change in temperature of the land and the evaporation of water and change humidity of the air. chemical change is the ripening of the orange</span>
Answer:
![\large \boxed{79 \, \%}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B79%20%5C%2C%20%5C%25%7D)
Explanation:
I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.
We have two conditions:
(1) Mass of glucose + mass of sucrose = 1.10 g
(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm
Let g = mass of glucose
and s = mass of sucrose. Then
g/180.16 = moles of glucose, and
s/342.30 = moles of sucrose. Also,
g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and
s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.
1. Set up the osmotic pressure condition
Π = cRT, so
![\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CPi_%7B%5Ctext%7Bg%7D%7D%20%2B%5CPi_%7B%5Ctext%7Bs%7D%7D%26%3D%26%5CPi_%7B%5Ctext%7Btot%7D%7D%5C%5C%5Cdfrac%7Bg%7D%7B450.4%7D%5Ctimes8.314%5Ctimes298%20%2B%20%5Cdfrac%7Bs%7D%7B855.8%7D%5Ctimes8.314%5Ctimes298%20%26%20%3D%20%26%203.78%5C%5C%5C%5C5.501g%20%2B%202.895s%20%26%20%3D%20%26%203.78%5C%5C%5Cend%7Barray%7D)
Now we can write the two simultaneous equations and solve for the masses.
2. Calculate the masses
![\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blrcl%7D%281%29%26%20g%20%2B%20m%20%26%20%3D%20%26%201.10%5C%5C%282%29%20%265.501g%20%2B2.895s%20%26%20%3D%20%26%203.78%5C%5C%283%29%20%26%20m%20%26%20%3D%20%261.10%20-%20g%5C%5C%265.501g%20%2B%202.895%281.10%20-%20g%29%20%26%20%3D%20%26%203.78%5C%5C%262.606g%20%2B%203.185%20%26%20%3D%20%26%203.78%5C%5C%20%262.606g%20%26%20%3D%20%26%200.595%5C%5C%284%29%20%20%26%20g%20%26%20%3D%20%26%20%5Cmathbf%7B0.229%7D%5C%5C%260.229%20%2B%20s%20%26%20%3D%20%26%201.10%5C%5C%26%20s%20%26%20%3D%20%26%20%5Cmathbf%7B0.871%7D%5C%5C%5Cend%7Barray%7D)
We have 0.229 g of glucose and 0.871 g of sucrose.
3. Calculate the mass percent of sucrose
![\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%7D%20%3D%20%5Cdfrac%7B%5Ctext%7BMass%20of%20component%7D%7D%7B%5Ctext%7BTotal%20mass%7D%7D%20%5Ctimes%20%5C%2C%20100%5C%25%5C%5C%5C%5C%5Ctext%7BPercent%20sucrose%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.871%20g%7D%7D%7B%5Ctext%7B1.10%20g%7D%7D%20%5Ctimes%20%5C%2C%20100%5C%25%20%3D%2079%20%5C%2C%20%5C%25%5C%5C%5C%5C%5Ctext%7BThe%20mixture%20is%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B79%20%5C%2C%20%5C%25%7D%7D%24%20sucrose%7D)
I am not sure plz show me the question