Since the sample size is below 30, in this case we use
the t statistic. The formula for t score is:
t = (x – u) / (σ / sqrt n)
where,
x = the level l = unknown
u = sample mean = 120 mg / dl
σ = standard deviation = 20 mg / dl
n = sample size or number of results = 5
Using the standard distribution tables for t, we can find
the value of t given the probability (P = 0.15) and degrees of freedom (DOF).
t = 1.036
Going back to the
formula for t score:
1.036 = (x – 120)
/ (20 / sqrt 5)
x = 129.27 mg /
dl = l
Answer:
a) Black
b) Non-magnetic
c) No reaction with carbon disulphide.
d) i don't know sry
But i had to go through my 6th grade notes for this .____.
Answer:
- a) 2N₂O(g) → 2N₂(g) + O₂(g)
Explanation:
Arrange the equations in the proper way for better understanding.
T<em>he reaction between nitrogen and oxygen is given below:</em>
<em />
- <em>2N₂(g) + O₂(g) → 2N₂O(g)</em>
<em />
<em>We therefore know that which of the following reactions can also occur?</em>
<em />
- <em>a) 2N₂O(g) → 2N₂(g) + O₂(g)</em>
- <em>b) N₂(g) + 2O₂(g) → 2NO₂(g)</em>
- <em>c) 2NO₂(g) → N₂(g) + 2O₂(g)</em>
- <em>d) None of the Above</em>
<h2>Solution</h2>
Notice that the first equation, a) 2N₂O(g) → 2N₂(g) + O₂(g), is the reverse of the original equation, 2N₂(g) + O₂(g) → 2N₂O(g).
The reactions in gaseous phase are reversible reactions that can be driven to one or other direction by modifying the conditions of temperature or pressure.
Thus, the equilibrium equation would be:
Which shows that both the forward and the reverse reactions occur.
Whether one or the other are favored would depend on the temperature and pressure: high temperatures would favor the reaction that consumes more heat (the endothermic reaction) and high pressures would favor the reaction that consumes more moles.
Thus, by knowing that one of the reactions can occur you can conclude that the reverse reaction can also occur.
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
