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Answer:
A. co dominace
incomplete dominance is when both genes show, this would result in the chicken being a black and white color
Explanation:
13.8 moles of carbon = 12 g/mol*13.8 or 165.6 grams
165.6 grams/ 2.26 g/mL = 73.27434 or 73.3 mL
I hope this helps!!!!
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation
Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
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