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3 years ago

which of these stars is least likely to be categorized as a supergiant? A. Betelgeuse B. Pollux C. Sirius A D. Sun Reset Submit

Chemistry

1 answer:

Vesna [10]3 years ago

5 0

Answer: The correct answer is option D.

Explanation: Super-giant stars are the stars which are greater than Sun. They have a mass hundred time greater than Sun and can be thousand times greater than Sun.

The masses of the stars are represented in Solar masses which is the mass of the Sun.

Mass of Betelgeuse = 20 Solar masses

Mass of Pollux = 1.7 Solar masses

Mass of Sirius = 2.02 Solar masses

Mass of Sun = 1 Solar mass.

As, the mass of Sun is the least from the given stars. Hence, it is least considered as a super-giant.

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3 years ago

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask

tangare [24]

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M$

0.18106 M is the molarity of the resulting solution.

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

$Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L$

0.823 Molar is the molarity of the solution.

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3 years ago