A patient who is prescribed a dose inhaler will find that it must be filled with a) medicine in powder form only. Works with lower (not upper) respiratory diseases only. Full of medicine used to give a fixed amount of medicine per oral inhalation. d) Medication in the form of a spray only.
NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
Answer:
Energy sources do not have 100% efficiency because <em>the processes of energy conversion to usable forms involves energy losses. </em>
Some have lower efficiencies due to; <u>energy losses in form of heat</u> during conversion, <u>poor technology applied during conversion</u> of energy and<u> lack of desire equipment</u> to use in the energy conversion system.
Explanation:
The desired form of energy for use is derived from conversion of energy from the source using an energy converter into another form which is usable. The efficiency of the energy converter is calculated as;
л = output energy/input energy
The efficiency of energy is limited to the cost of equipment required for conversion from energy source by the energy converter to a form which is usable. Additionally, because energy sources are scarce, the technology to use in energy conversion is a factor affecting energy efficiency in that high efficiency will require advanced technology with better equipment leading higher costs of that energy form. when heat losses are involved during energy conversion, efficiency lowers, thus its better if such losses are used as energy input in another system.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
