Answer:
Option A, C, D
Explanation:
A reversible adiabatic process is an isentropic process as entropy remains constant. Thus,
Entropy change (Delta S) = integral of (Q_rev)/T
Thus, option A and C are correct
For a reversible process, entropy generation is a positive quantity or zero. hence, statement B is not true
In an Adiabatic processes, entropy do not decrease. Hence, option D is also correct
It will spread out making the sound travel faster then it usually does in the air(i think) let me know if you need help with anything else
Answer:
Option B. 2Mg(s) + O2 (g) —> 2MgO (s)
Explanation:
From the question given above,
We were told that:
2 solid Mg atoms bond with O2 gas to produce solid MgO.
This can be represented by an equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
Next, we shall balance the above equation as follow:
2Mg(s) + O2 (g) —> MgO (s)
There are 2 atoms of Mg on the left side and 1 atom on the right side. It can be balance by putting 2 in front of MgO as shown below:
2Mg(s) + O2 (g) —> 2MgO (s)
Now, the equation is balanced.
Answer:
=<em><u> 0.42 moles of CO2 </u></em>
Explanation:
From Avogadro's constant
6.02×10^23 molecules are in 1 mole of CO2
2.54×10^23 molecules will be in
=[(2.54×10^23) ÷ (6.02×10^23)]
= 0.42 moles of CO2
Yes it is used , hope this helps
