Answer:
No matter how well how much you accomplish, or how much good you put into the world, you will always have haters. An unfortunate aspect of life is that some people hate when others do well.It’s just a way of life.some people just are jealous but dont let people brake you down.
Explanation:
hope this helps
Answer:
fundamental frequency in helium = 729.8 Hz
Explanation:
Fundamental frequency of an ope tube/pipe = v/2L
where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L is length of the pipe
To find the length of the pipe,
frequency = velocity of sound / 2L
272 = 340 / 2 L
L = 0.625 m
If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.
Also, the velocity of sound is inversely proportional to square root of molar mass of gas; v ∝ 1/√m
v₁/v₂ = √m₂/m₁
v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium
340 / v = √4 / 28.8
v₂ = 340 / 0. 3727
v₂ = 912.26 m /s
fundamental frequency in helium = v₂ / 2L
fundamental frequency in helium = 912.26 / (2 x 0.625)
fundamental frequency in helium = 729.8 Hz
Answer:
wave length=speed/frequency
=345/229
Answer:
B) ) –1615.1 kJ mol^–1
Explanation:
since
SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1
the enhalpy of reaction will be
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants
therefore
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]
4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol
∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol
therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol
