Eye
Explanation:
Vitreous humor is found in the human eye and other animals.
It is a gel that fills the space between the lens and retina of the eye. This matter helps to keep the shape of the eye in place by maintaining a constant pressure in the eye.
It is typically made up of water, gelatinous and transparent.
Learn more:
Human eye brainly.com/question/8032392
Color in the eye brainly.com/question/9434044
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The <span>molar concentration of the crystal violet solution is more concentrated than that of the sodium hydroxide solution. It is because the crystal violet solution has more solute in it compared to the sodium hydroxide.</span>
The average weight of an atom of an element, formerly based on the
weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the
weight of an oxygen atom, but after 1961 based on 1/12 the weight of the
carbon-12 atom.
Answer:
The concentration the student should write down in her lab is 2.2 mol/L
Explanation:
Atomic mass of the elements are:
Na: 22.989 u
S: 32.065 u
O: 15.999 u
Molar mass of sodium thiosulfate, Na2S2O3 = (2*22.989 + 2*32.065 + 3*15.999) g/mol = 158.105 g/mol.
Mass of Na2S2O3 taken = (19.440 - 2.2) g = 17.240 g.
For mole(s) of Na2S2O3 = (mass taken)/(molar mass)
= (17.240 g)/(158.105 g/mol) = 0.1090 mole.
Volume of the solution = 50.29 mL = (50.29 mL)*(1 L)/(1000 mL)
= 0.05029 L.
To find the molar concentration of the sodium thiosulfate solution prepared we use the formula:
= (moles of sodium thiosulfate)/(volume of solution in L)
= (0.1090 mole)/(0.05029 L)
= 2.1674 mol/L
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:
Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n
According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:
Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)
