C2H6O + O2 ---> C2H4O2 + H2O
using the molar masses:-
24+ 6 + 16 g of C2H6O produces 24 + 4 + 32 g C2H4O2 (theoretical)
46 g produces 60g
60 g C2H4O2 is produced from 46g C2H6O
1g . .................................46/60 g
700g ................................. (46/60) * 700 Theoretically
But as the yield is only 7.5%
the required amount is ((46/60) * 700 ) / 0.075 = 7155.56 g
= 7.156 kg to nearest gram. Answer
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)
It's C, the amount of protons in the nucleus of each atom :)
Average velocity is the change in distance divided by the time it takes to travel that distance. so the first one.
Answer:
They are all located in the same horizontal row or period.
Explanation:
In the periodic table the elements are arranged according to their increasing atomic number.The elements those contain low atomic number are placed in the first row such as hydrogen and helium. similarly the radioactive elements those have high atomic number are placed in 7th or 8th row.
All the elements of a family in the peroidic table have common features are located in same horizontal row or period.For example fluorine,chlorine,bromine and iodine are placed in VII B class of periodic table as all of them are electronegative in nature.