Answer:
50%
Explanation:
% yields =(experimental yields / theoretical yields ) x100%
= 122.1÷ 245.6 ) x100% = 50%
<u>Answer:</u> The mass percent of ethylene glycol in solution is 14.2 %
<u>Explanation:</u>
We are given:
Molarity of ethylene glycol solution = 2.45 M
This means that 2.45 moles of ethylene glycol is present in 1 L or 1000 mL of solution.
- To calculate the mass of solution, we use the equation:
![\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}](https://tex.z-dn.net/?f=%5Ctext%7BDensity%20of%20substance%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20substance%7D%7D%7B%5Ctext%7BVolume%20of%20substance%7D%7D)
Density of solution = 1.07 g/mL
Volume of solution = 1000 mL
Putting values in above equation, we get:
![1.07g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.07g/mL\times 1000mL)=1070g](https://tex.z-dn.net/?f=1.07g%2FmL%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solution%7D%7D%7B1000mL%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20solution%7D%3D%281.07g%2FmL%5Ctimes%201000mL%29%3D1070g)
- To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Moles of ethylene glycol = 2.45 moles
Molar mass of ethylene glycol = 62 g/mol
Putting values in above equation, we get:
![2.45mol=\frac{\text{Mass of ethylene glycol}}{62g/mol}\\\\\text{Mass of ethylene glycol}=(2.45mol\times 62g/mol)=151.9g](https://tex.z-dn.net/?f=2.45mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20ethylene%20glycol%7D%7D%7B62g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20ethylene%20glycol%7D%3D%282.45mol%5Ctimes%2062g%2Fmol%29%3D151.9g)
- To calculate the mass percentage of ethylene glycol in solution, we use the equation:
![\text{Mass percent of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20ethylene%20glycol%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20ethylene%20glycol%7D%7D%7B%5Ctext%7BMass%20of%20solution%7D%7D%5Ctimes%20100)
Mass of solution = 1070 g
Mass of ethylene glycol = 151.9 g
Putting values in above equation, we get:
![\text{Mass percent of ethylene glycol}=\frac{151.9g}{1070g}\times 100=14.2\%](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20ethylene%20glycol%7D%3D%5Cfrac%7B151.9g%7D%7B1070g%7D%5Ctimes%20100%3D14.2%5C%25)
Hence, the mass percent of ethylene glycol in solution is 14.2 %
Hi! In many exam questions, they usually inform you on which number to use for the acceleration due to gravity. Its actual value is 9.80214 ms^(-2) however some exam questions ask you to use 9.8 or even 10...so I did both just for the extra help. Hope this helps!
Answer:
the precise location of electrons at a specific time is unknown
Explanation:
The electron cloud model states that an electron’s position is uncertain, and does not revolve around the nucleus along a set course, as proposed by Bohr.