Answer:
The answer is 0.046 mol.
Explanation:
By looking at the balanced equation, you can form a ratio of lithium chloride and lithium bromide using the coefficient value :
ratio of lithium bromide <u>(</u>2LiBr)
= 2
ratio of lithium chloride (2LiCl)
= 2
So the ratio is 2 : 2 then simplify into 1 : 1 . Which means that 1 mol of lithium bromide is equal to 1 mole of lithium chloride.
In this case, 0.046 mol of lithium bromide will form <u>0</u><u>.</u><u>0</u><u>4</u><u>6</u><u> </u><u>m</u><u>o</u><u>l</u> of lithium chloride.
1.85 quarts can fit into a 1.75 liter bottle
Answer: the scientist should ensure that the volume is measured in dm3(or L), the pressure in atm, the temperature in Kelvin and use the gas constant 0.082atm.dm3.K^-1.mol^-1
Answer:
4600s
Explanation:

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:
![-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3Dk%5BB%5D%20-%20-%20-%20%20-%5Cfrac%7Bd%5BB%5D%7D%7B%5BB%5D%7D%3Dk%2Adt)
If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.
PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.
![-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3Dk%2Adt)
Integrating we get:
![\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ep%20%5C%2C-%5Cfrac%7Bd%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%7D%7BP%28N_%7B2%7DO_%7B5%7D%29%7D%3D%5Cint%5Climits%5E%20t%20k%2Adt)
![-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})](https://tex.z-dn.net/?f=-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%3Dk%28t_%7B2%7D-t_%7B1%7D%29)
Clearing for t2:
![\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%28ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D-ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%29%5D%29%7D%7Bk%7D%2Bt_%7B1%7D%3Dt_%7B2%7D)
![ln[P(N_{2}O_{5})]=ln(650)=6.4769](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29%5D%3Dln%28650%29%3D6.4769)
![ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333](https://tex.z-dn.net/?f=ln%5BP%28N_%7B2%7DO_%7B5%7D%29_%7Bo%7D%5D%3Dln%28760%29%3D6.6333)

Answer: D) 36.0 km/hr, downstream
Explanation:
For downstream motion of the boat, the actual velocity of the boat is the sum of velocity of the water current and the velocity of the boat due to pushed by wind.
Velocity of water current, v = 15 km/h
Velocity of the boat going downstream, u = 21 km/h
Actual velocity of the boat = v'
v' = v + u
⇒v' = 15 km/h + 21 km/h
⇒u = 21 km/h +15 km/h = 36.0 km/h downstream
Thus, the correct answer is option D.