Answer:
The system can operate on a set level of parts per million (ppm) chlorine, where the set level is automatically maintained.
Answer:
m = 1.3 Kg
Explanation:
according to Specific Heat Formula :
Q = mc∆T
when Q is the heat energy in Joules
m is the mass of water
C is the specific heat its unit J/kg∙K
∆T change in temperature (the final t - the initial t)
by substitution:
Q = 190000 g * 4.184 J/g∘C *78
Q = 62006880 J
when The combustion of 1.0 lb of oil provides 2.2×107J
So lb of oil when Q = 62006880 J = 2.8 Ib
and when 1lb=454g
So 2.8 Ib = 1271.2 g = 1.3 Kg
Answer:
ΔH°r = -1562 kJ
Explanation:
Let's consider the following combustion.
C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)
We can calculate the standard heat of reaction (ΔH°r) using the following expression:
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
where,
ni are the moles of reactants and products
ΔH°f(i) are the standard heats of formation of reactants and products
The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0
ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)
ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]
ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]
ΔH°r = -1562 kJ
Mother answers are Neon and Argon
