Answer: 7.79 grams of ethanol were put into the beaker.
Explanation:
To calculate the mass of ethanol, we use the equation:
Density of ethanol = 0.779 g/mL
Volume of water = 10.00 mL
Putting values in above equation, we get:
Thus 7.79 grams of ethanol were put into the beaker.
Assuming that the reactants are:
(NH4)2SO4 (aq) + Ba(NO3)2 (aq)
and the products are:
BaSO4 (s) + 2NH4NO3 (aq),
then you will have to determine which product is insoluble. You should have access to solubility rules to help you determine this.
According to the solubility rules, the following elements are considered insoluble when paired with SO4:
Sr^2+, Ba^2+, Pb^2+, Ag^2+, and Ca^2+
Therefore, the precipitate will be BaSO4 (s).
Moles of water atoms = mass/molecular weight = 105/18 = 5.83 mol. Number of moles of hydrogen in water = 2 x moles of water = 11.66. Number of H atoms in water = moles of hydrogen x 6.02 x 10^23 = 7.019 x 10^24 ~ 7.02 x 10^24 atoms. Hope this helps.
Answer:
Explanation:
mass % of C = 0.27/0.45*100 = 60%
mass % of H = 0.02/0.45*100 = 4.4%
mass % of O = 0.16/0.45*100 = 35.6%
Total = 60%+4.4%+ 35.6% = 100%
