So in order for us to know the percentage of sugar present in a 12.00 g of milk chocolate, what we are going to do is that, we just have to divide 8 by 12 and multiply in by 100 and we get 66.67. Therefore, the percentage of sugar present in 12.00 g of milk chocolate bar is 66.67%. Hope this answers your question. Have a great day!
Answer:
Explanation:
The result will be affected.
The mass of KHP weighed out was used to calculate the moles of KHP weighed out (moles = mass/molar mass).
Not all the sample is actually KHP if the KHP is a little moist, so when mass was used to determine the moles of KHP, a higher number of moles than what is actually present would be obtained (because some of that mass was not KHP but it was assumed to be so. Therefore, there is actually a less present number of moles than the certain number that was thought of.
During the titration, NaOH reacts in a 1:1 ratio with KHP. So it was determined that there was the same number of moles of NaOH was the volume used as there were KHP in the mass that was weighed out. Since there was an overestimation in the moles of KHP, then there also would be an overestimation in the number of moles of NaOH.
Thus, NaOH will appear at a higher concentration than it actually is.
Answer: The standard enthalpy of formation of 
is -252.1 kJ/mol.
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactantss}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactantss%7D%5D)
Putting the values we get :
![\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B2%5Ctimes%20H_f%7BFe%7D%2B3%5Ctimes%20H_f%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f%7BFe_2O_3%7D%2B3%5Ctimes%20H_f%7BH_2%7D%5D)
![67.9kJ=[(2\times 0)+(3\times H_f{H_2O})]-[(1\times -824.2kJ/mol)+3\times 0kJ/mol)]](https://tex.z-dn.net/?f=67.9kJ%3D%5B%282%5Ctimes%200%29%2B%283%5Ctimes%20H_f%7BH_2O%7D%29%5D-%5B%281%5Ctimes%20-824.2kJ%2Fmol%29%2B3%5Ctimes%200kJ%2Fmol%29%5D)
Thus standard enthalpy of formation of is -252.1 kJ/mol.
The answer is Noble gas because they have a full octet
False, our tongue and nose work together
