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daser333 [38]
3 years ago
13

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
What is the boiling point of a solution containing 203 g of ethylene glycol (C2H6O2) and 1035 g of water? (Kb for water is 0.52
Andru [333]

Answer:

101,37°C

Explanation:

Boiling point elevation is one of the colligative properties of matter. The formula is:

ΔT = kb×m <em>(1)</em>

Where:

ΔT is change in boiling point: (X-100°C) -X is the boiling point of the solution-

kb is ebulloscopic constant (0,52°C/m)

And m is molality of solution (mol of ethylene glycol / kg of solution). Moles of ethylene glycol (MW: 62,07g/mol):

203g × (1mol /62,07g) = <em>3,27moles of ethlyene glycol</em>

<em />

Molality is: 3,27moles of ethlyene glycol / (1,035kg + 0,203kg) = 2,64m

Replacing these values in (1):

X - 100°C = 0,52°C/m×2,64m

X - 100°C = 1,37°C

<em>X = 101,37°C</em>

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I hope it helps!

7 0
3 years ago
How does ionization energy change as you
kykrilka [37]

Answer:

How does the energy required to remove an electron from an atom change as you move left to right in Period 4 from potassium through iron? ... A greater nuclear charge pulls the electrons closer to the nucleus, decreasing the atomic radius.

3 0
2 years ago
Consider the reaction 2CO * O2 —&gt; 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c
Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

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3 years ago
What do we mean by basic MgO?
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3 years ago
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