Answer:
Name Formula
nitrite ion NO2−
permanganate ion MnO4−
phosphate ion PO43−
hydrogen phosphate ion HPO42−
Explanation:
I’m pretty sure the answer is D :)
279 g * (1 mol/180.559g glucose) * (2 mol ethanol/1 mol glucose) * (46.068g ethanol/1mol) =
142 g ethanol produced
The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.
Molarity is mol HNO3 / L of solution. This is our aim
The given percentage is 0.68 g HNO3/ g solution
multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain
0.016 mol/ mL or 16.23 mol/ L (M)
