Answer:
red supergiants is the answer
Answer: 0.745 g of 
will be produced from 1.08 g of sodium sulfate
Explanation:
To calculate the moles :
is the limiting reagent as it limits the formation of product and 
is the excess reagent.
According to stoichiometry :
3 moles of produce = 3 moles of
Thus 0.0076 moles of will require=
of
Mass of 
Thus 0.745 g of will be produced from 1.08 g of sodium sulfate
0.3147 concentration (in moles/l) of a saline (NaCl) solution will provide an isotonic eyedrop solution.
Isotonic eye drops
Because it might result in eye discomfort or tissue damage if it is not maintained, isotonicity is regarded as a crucial component of ophthalmic medicines. A few drops of blood are mixed with the test preparation before being examined and judged under a microscope at a magnification of 40. Isotonic solutions are those that have the same amount of water and other solutes in them as the cytoplasm of a cell. Since there is no net gain or loss of water, placing cells in an isotonic solution will not cause them to either shrink or swell.
We can calculate the osmotic pressure exerted by a solution using the following expression.
π = M . R . T
where,
π is the osmotic pressure
M is the molar concentration of the solution
R is the ideal gas constant
T is the absolute temperature
The absolute temperature is 37 + 273 = 310 K
π = M . R . T
8 = (X mol/L) . (0.082atm.L/mol.K) . 310 K = 0.3147 mol/L
To learn more about osmotic pressure refer:
brainly.com/question/5041899
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<u>Answer:</u> The percent composition by mass of hydrogen in given compound is 6.33 %
<u>Explanation:</u>
We are given:
A chemical compound having chemical formula of 
It is made up by the combination of 1 nitrogen atom, 5 hydrogen atoms, 1 carbon atom and 3 oxygen atoms
To calculate the percentage composition by mass of hydrogen in the compound, we use the equation:
Mass of compound = ![[(1\times 14)+(5\times 1)+(1\times 12)+(3\times 16)]=79g/mol](https://tex.z-dn.net/?f=%5B%281%5Ctimes%2014%29%2B%285%5Ctimes%201%29%2B%281%5Ctimes%2012%29%2B%283%5Ctimes%2016%29%5D%3D79g%2Fmol)
Mass of hydrogen = 
Putting values in above equation, we get:
Hence, the percent composition by mass of hydrogen in given compound is 6.33 %
