Answer:
The correct statements are:
The rate of disappearance of B is twice the rate of appearance of C.
Explanation:
Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.
3A + 2B → C + 2D
Rate of the reaction:
![R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=R%3D-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
The rate of disappearance of B is twice the rate of appearance of C.
![\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}](https://tex.z-dn.net/?f=2%5Ctimes%20%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B1%7D%5Ctimes%20%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
Answer:
About 1.301 atm
Explanation:
The formula that you should is PV=nRT, where P stands for pressure, V stands for volume, n stands for the number of moles, R stands for the universal gas constant, and T stands for temperature in Kelvin. Since the volume, number of moles, and universal gas constant don't change, you don't need to worry about them.
1.07V=393nR
PV=498nR
P=1.301 atm. Hope this helps!
Answer:
The volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L
Explanation:
Given data:
Number of moles of HF = 6.62×10⁻³ mol
Volume of HF in litter at STP = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Standard temperature = 273 K
Standard pressure = 1 atm
Now we will put the values in formula.
1 atm × V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K
V = 6.62×10⁻³mol ×0.0821 atm.L/ mol.K × 273 K / 1 atm
V = 148.38×10⁻³ L
Thus, the volume of 6.62×10⁻³moles of HF at STP is 148.38×10⁻³ L.
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