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timurjin [86]
3 years ago
11

When an element exist in nature by itself, it must have a charge of

Chemistry
1 answer:
pochemuha3 years ago
3 0
It must have a charge of zero

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∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821
Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

<h3>4) 0.26 atm </h3><h3 />
6 0
2 years ago
In which example does entropy increase?
nikklg [1K]

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

<u>entropy increases as the randomness increase , </u>

and ,

<u>entropy decrease as the randomness decreases . </u>

I. A piece of fruit freezes in liquid nitrogen.

randomness decreases , and so entropy decrease .

II. Librarians put library books back onto shelves at the end of the day.

randomness decreases , and so entropy decrease .

III. Liquid nitrogen boils in a beaker at room temperature.

change in state from less random liquid to gas , randomness increases , hence entropy increases

IV. Dry ice sublimes .

change in state from less random solid to gas , randomness increases , hence entropy increases .

V. A balloon collapses as its contents cool in liquid nitrogen .

cooling process , reduces the randomness , hence , entropy decreases .

5 0
3 years ago
Hibernation is _______. A. An example of an abiotic cycle b. Triggered by changes in the weather c. Caused only by biotic factor
natta225 [31]

The correct answer is option b, that is, triggered by changes in the weather.  

Hibernation refers to a state of lower metabolism and inactivity usually witnessed in endotherms. This condition is featured by a slow heart rate, slow breathing, low metabolic rate, and low body temperature. It assists the majority of the endotherms to thrive in the cold seasons. For example, hibernation of a bear in winters. Therefore, weather change can stimulate hibernation.  


6 0
3 years ago
An atom of uranium-238 undergoes radioactive decay to form an atom of thorium-234. Which type of nuclear decay has occurred?
WINSTONCH [101]

Answer:-

Alpha decay

Explanation:-

Uranium 238 has atomic number 92 and mass number 238.

Thorium 234 has atomic number 90 and mass number 234.

So, the change in atomic number as uranium 238 disintegrates into thorium234 = 92 – 90 = 2

So, the change in mass number as uranium 238 disintegrates into thorium234= 238 – 234 = 4

We know that when an alpha particle is emitted, the mass number decreases by 4 and the atomic number decreases by 2.

So when an atom of uranium 238 undergoes radioactive decay to form an atom of thorium-234, alpha decay has occurred.

5 0
2 years ago
Question 43 please. i’m not sure if i got the right answer
hram777 [196]

a) The reaction is exothermic since the overall enthalpy change is negative. this means that the system has lost energy to the environment, namely, the apparatus and due to drought.

b) We first calculate the number of moles in 3.55 grams of magnesium.

number of moles= mass/ atomic mass

                            =3.55/24

                            =0.1479 moles(to 4sf)

now, if 2 moles of magnesium give -1204kJ

        How much energy is given by 0.1479 moles

       = (0.1479×-1204kJ)

        =-89.0358kJ (don't forget the negative sign)

c) two molesof MgO produces -1204kJ of energy

then -234kJ will be produced by

=(-234kJ×2moles)/1204kJ

=0.3887moles

one mole of MgO weighs 24+16=40

therefore the mass produced is 0.3887moles×40=15.548grams

(d) we first find the number of moles of MgO in 40.3 grams

number of moles=mass/RFM

=40.3g/40= 1.0075moles

if 2 moles of MgO give 1204 kJ then decomposing 1.0075 moles requires

(1.0075 moles×1204kJ)/2=606.515kJ

4 0
3 years ago
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