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Veseljchak [2.6K]
3 years ago
7

Which of the following changes will always be true for an exothermic process?

Mathematics
2 answers:
Ivenika [448]3 years ago
6 0
Given that I no not see any explanation in the previos answer, I think you can benefit of this.

Exothermic refers only to the fact that the reaction releases energy in the form of heat when occurs. So, the enthalpy is higher at start and lower and the end.

Seen in this way, you can deduct that the change in enthalpy is negative, which is what the third option says. 
Tamiku [17]3 years ago
5 0
3rd option is right.

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A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par
lyudmila [28]

Answer:

a(\frac{1}{5e})=5e

Step-by-step explanation:

we are given equation for position function as

s(t)=tln(5t)

Since, we have to find acceleration

For finding acceleration , we will find second derivative

s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)

=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t

=1\cdot \ln \left(5t\right)+\frac{1}{t}t

s'(t)=\ln \left(5t\right)+1

now, we can find derivative again

s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)

=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)

=\frac{1}{t}+0

a(t)=\frac{1}{t}

Firstly, we will set velocity =0

and then we can solve for t

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we get

t=\frac{1}{5e}

now, we can plug that into acceleration

and we get

a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}

a(\frac{1}{5e})=5e


5 0
3 years ago
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