A student is asked to standardize a solution of potassium hydroxide. He weighs out 0.996 g potassium hydrogen phthalate (KHC2H404, treat this as a monoprotic acid).
It requires 22.0 mL of potassium hydroxide to reach the endpoint.
A. What is the molarity of the potassium hydroxide solution? M
This potassium hydroxide solution is then used to titrate an unknown solution of nitric acid,
B. If 14.9 mL of the potassium hydroxide solution is required to neutralize 14.9 mL of nitric acid, what is the molarity of the nitric acid solution?
M

Nanotechnology, the field of trying to build ultrasmall structures one atom at a time, has progressed in recent years. one potential application of nanotechnology is the construction of artificial cells. the simplest cells would probably mimic red blood cells, the body's oxygen transporters. for example, nanocontainers, perhaps constructed of carbon, could be pumped full of oxygen and injected into a person's bloodstream. if the person needed additional oxygen-due to a heart attack perhaps, or for the purpose of space travel-these containers could slowly release oxygen into the blood, allowing tissues that would otherwise die to remain alive. suppose that the nanocontainers were cubic and had an edge length of 24 nanometers. part a part complete what is the volume of one nanocontainer? (ignore the thickness of the nanocontainer's wall.) express your answer using two significant figures. v v = 1.4ă—10â’20 l previous answers correct significant figures feedback: your answer 1.3824â‹…10â’20 = 1.382ă—10â’20 l was either rounded differently or used a different number of significant figures than required for this part. if you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer. part b suppose that each nanocontainer could contain pure oxygen pressurized to a density of 81 g/l . how many grams of oxygen could be contained by each nanocontainer?

At 450 mm hg a gas has a volume of 760 l, what is its volume at standard pressure

Given the following at 25c calculate delta hf for hcn (g) at 25c. 2nh3 (g) +3o2 (g) + 2ch4 (g) > 2hcn (g) + 6h2o (g) delta h rxn= -870.8 kj. delta hf=-80.3 kj/mol for nh3 (g), -74.6 kj/mol for ch4, and -241.8 kj/mol for h2o (g)