Explanation:
ΔS denotes change in entropy. Entropy is the degree of disorderliness of a system. An increasing ΔS would mean that the entropy of the products is greater than the entropy of the reactants.
Generally the trend of entropy with state of matter is given as;
Gas > Liquid > Solid (Increasing Entropy)
a. H 2O(g) → H 2O(l)
In this reaction, there is a decreasing ΔS, since we are moving from gas to liquid. This would be the last.
b. 2NO(g) → N 2(g) + O 2(g)
There is an increase in gaseous products. This is definitely an increasing ΔS reaction. This would be the first on the list.
C. MgCO 3(s) → MgO(s) + CO2(g)
The products formed are in solid and gaseous state. There is increasing ΔS but not up to reaction 2.
The order is given as;
b. 2NO(g) → N 2(g) + O 2(g)
C. MgCO 3(s) → MgO(s) + CO2(g)
a. H 2O(g) → H 2O(l)
Answer:
Each student will need;
1 red jelly bean, 1 white jelly bean, 1 black jelly bean and 3 red jelly beans.
Explanation:
Sodium bicarbonate molecule, NaHCO3, or baking soda is composed of the following:
1 atom of sodium, Na;
1 atom of hydrogen, H;
1 atom of carbon, C, and
3 atoms of oxygen.
For each of the models to be built by the two students, these atoms are to be represented accordingly.
Since Red jelly beans represent sodium atoms (Na), white jelly beans represent hydrogen atoms (H), black jelly beans represent carbon atoms (C), and blue jelly beans represent oxygen atoms (O), each student will need;
1 red jelly bean, 1 white jelly bean, 1 black jelly bean and 3 red jelly beans.
A sample model is found in the attachment below:
As we know that Molarity is given as,
M = moles / V
Solving for V,
V = moles / M ------------------(1)
Also, moles is equal to,
moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
V = (mass / M.mass) / M
Putting values,
V = (45 g / 164 g/mol) / 1.3 mol/dm³
V = 0.21 dm³
Answer:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Explanation:
Using Boyle's law
Given ,
V₁ = 3.6 L
V₂ = ?
P₁ = 1.0 atm
P₂ = 13.3 atm (From correct source)
Using above equation as:
The volume that this same amount of air will occupy in his lungs when he reaches a depth of 124 m is - 0.27 L.
Answer: The workdone W = 505J
Explanation:
Applying the pressure-volume relationship
W= - PΔV
Where negative sign indicates the power is being delivered to the surrounding
W = - 1.0atm * ( 5.88 - 0.9)L
= - 1.0atm * (4.98)
W = -4.98 atmL
Converting to Joules
1atmL = 101.325J
-4.98atmL = x joules.
Work done in J = -4.98 * 101.325
W= -505J
Therefore the workdone is -505J
