My Apoligies for not being sooner.
Answer: The closer the particles get or the further apart they get, the greater the sound's amplitude. Sound amplitude causes a sound's loudness and intensity. The bigger the amplitude is, the louder and more intense the sound. - (This is copied)
General Answer : Its a change in sound, bassically the bigger the waves are, the louder the sound is.
CxHy + O2 --> x CO2 + y/2 H2O
Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound
Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound
Find the mass of C and H in the compound:
.430mol x 12 = 5.16 g C
.166mol x 1g = .166g H
When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.
Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).
In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).
Answer : The value of work done by an ideal gas is, 37.9 J
Explanation :
Formula used :
Expansion work = External pressure of gas × Volume of gas
Expansion work = 1.50 atm × 0.25 L
Expansion work = 0.375 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 0.375 × 101.3 = 37.9 J
Therefore, the value of work done by an ideal gas is, 37.9 J
Answer:
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
Explanation:
Let the mass of the first solution be x and second solution be y.
Amount solution required = 1250 kg
x + y = 1250 kg....[1]
Percentage of ethanol in required solution = 12% of 1250 kg
Percentage of ethanol in solution-1 = 5% of x
Percentage of ethanol in required solution = 25% of y
5% of x + 25% of y =12% of 1250 kg
x + 5y = 3000 kg...[2]
Solving [1] and [2] we :
x = 437.5 kg , y = 812.5 kg
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
