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jek_recluse [69]
2 years ago
9

What is helium an example of? element compound atomic number subatomic particle​

Chemistry
2 answers:
fgiga [73]2 years ago
4 0

Answer:

atomic number

Explanation:

hope this helps

Hatshy [7]2 years ago
3 0
I believe its element.
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Ice cubes are placed on a table in a warm room. What happens to the ice particles?
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B

Explanation:

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How many inner, outer, and valence electrons are present in an atom of each of the following elements?
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Answer:

(a) O =  Valance Electrons (6), Inner electrons (8)

(b) Sn =  Valance Electrons (2), Inner electrons (36)

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The property of a substance stays ______________ no matter how the substance is used.
Irina18 [472]

Any substance changes to another substance that means the change of the physical property. Like water () has different state which changes as the temperature changes. It remain as liquid in the room temperature, in solid form at or below 0°C and vapor phase on or above 100°C. But in all the stage or phase of the substance the composition of the water i.e.  remains. Thus the chemical property remains fixed when a substance change to other substance.  

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3 years ago
Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
Ghella [55]

Answer:

(a) ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = -2.881 J/K; total change in entropy = 0

(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

(c) ΔS_{sys}  = 0 ; ΔS_{sur}  = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

6 0
3 years ago
How would the amount of force needed to keep the kite in the air change if it were made out of a heavier material instead of a l
erik [133]
A: the amount of force would need to be greater.
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Read 2 more answers
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