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Dovator [93]
3 years ago
8

Which body changes occur after an individual has become an adult?

Chemistry
2 answers:
sweet-ann [11.9K]3 years ago
7 0
The answer is puberty.
morpeh [17]3 years ago
3 0
Puberty is what is correct I believe
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What should you be careful of when measuring the temperature of a liquid in a beaker
andrezito [222]

The precaution to be taken while measuring the temperature of a liquid in a beaker is applying proper heat balance and taking all the required precautions.

  • A beaker with an open top contains a sample of liquid. It exposes this sample to light.
  • That liquid absorbs the light energy, turning it into heat energy. As a result, the liquid becomes warmer and evaporation is accelerated. As a result, there is less liquid in the beaker.
  • Since it is well known that the surface temperature of a liquid, along with air movement above the liquid surface, is one of the dominant factors affecting evaporation, I want to measure the evaporation rate as a function of surface temperature.
  • This can be done by applying a heat balance.

Learn more about heat balance at:

brainly.com/question/1292905

#SPJ9

3 0
2 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
Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles O2 to the
jek_recluse [69]

Answer:

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Explanation:

Step 1: Data given

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate the mol ratio

For 3 moles O2 we'll have 4 moles Fe

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Option b says we have 2 mole Fe2O3 for each 4 moles Fe

This doesnt say anything about O2. So doesn't apply for this question.

Option C says we have 4 moles of Fe for each 2 moles Fe2O3

This is the same as option B, so doesn't apply for this question.

Option D says for each 3 moles of O2 we have 2 Fe2O3

This is true, but doesn't say anything about Fe so doesn't apply here.

4 0
3 years ago
Read 2 more answers
Please help? Liquid, Chemical Properties, Compounds or Products?​
Rainbow [258]

Answer:

They are compounds.

Explanation:

..........

3 0
3 years ago
Read 2 more answers
9.8 x 10​^-6 regular notation
topjm [15]

Answer:

0.0000098 should be the answer

Explanation:

3 0
3 years ago
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