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nikklg [1K]
3 years ago
5

Which characteristic is not used to classify stars? size color distance from Earth age

Chemistry
2 answers:
Westkost [7]3 years ago
8 0
According to the topic on stars, galaxies and the universe, the characteristics that are used to classify stars are color, temperature, size, elemental or compound composition, and brightness. Among these factors give, the characteristic not used to classify stars is distance from the earth. 
iren2701 [21]3 years ago
5 0

Answer:

Distance from earth

Explanation:

The star is classified based on following five basic properties:

1) brightness of star

2) color of star

3) surface temperature of star

4) size of star

5) mass of star

So based on this the only characteristics which is not used to classify stars is distance from earth as size, brightness, temperature are also related to age of stars.

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Which of the following is similar among all living organisms on Earth?
statuscvo [17]

A. cellular function

 hope this helps :)

5 0
3 years ago
Define density of a substance.
ozzi

Answer:

density, mass of a unit volume of a material substance. The formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre. ... Density can also be expressed as kilograms per cubic metre (in metre-kilogram-second or SI units).

Explanation:

4 0
3 years ago
Read 2 more answers
Learning calculations in chemistry
olasank [31]
What question do u have?
7 0
3 years ago
For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper
Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1\text{ and }V_1 are initial pressure and volume at initial temperature T_1.

P_2\text{ and }V_2 are final pressure and volume at initial temperature T_2.

We are given:

P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V

T_1=-25^oC=248.15 K

T _2 = 16^oC=289.15 K

Putting values in above equation, we get:

\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}

P_1=0.225 kPa

Hence, the new pressure will be 0.225  kPa.

4 0
4 years ago
When chromium loses two electrons, its configuration changes to
damaskus [11]

Electronic configuration of cromium is  

Cr-[Ar]4s¹3d⁵  

When cromium loses two electrons it becomes Cr⁺².  

So its electronic configuration becomes,  

Cr⁺²-[Ar]3d⁴  

One electron will go from 4s orbital and one electron will go from 3d orbital.

So the answer here is D. [Ar]3d⁴ -because after losing 2 electrons electronic configuration of cromium becomes  [Ar] 3d⁴.


3 0
3 years ago
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