N=5.0 mol
m=3.2 kg
C=n/m
C=5.0/3.2≈1.6 mol/kg
Answer:
the correct answer is option C. Na
Answer:
N and P
Explanation:
Anion:
When an atom gain the electrons anion is formed. The negative sign shows that atom gain electron because number of electron are greater than protons or we can say that negative charge becomes greater than positive charge.
Cation:
When atom lose electron cation is formed. The atom thus have positive charge because number of positive charge i.e protons are increased are greater than negative charge or electron.
In given problem N and phosphorus both can gain three electrons which means negative charge becomes greater that's why the extra electron gained by atoms are written as -3 and both form anion with charge -3.
while Al form cation with charge +3 Mg form cation with charge +2 and iodine and bromine both form anion with charge of -1.
Answer: The Gregorian calendar month, which is 1⁄12 of a tropical year, is about 30.44 days, while the cycle of lunar phases (the Moon's synodic period) repeats every 29.53 days on average. Therefore, the timing of the lunar phases shifts by an average of almost one day for each successive month.
Explanation: This is what GOOGLE says
hope it helps a little!!
Answer:
(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt
(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls
Explanation:
By definition, t<u>he reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:</u>
aX → bY (1)
![rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%3D%20-%5Cfrac%7B1%7D%7Ba%7D%20%5Cfrac%7B%5CDelta%5BX%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7Bb%7D%20%5Cfrac%7B%5CDelta%5BY%5D%7D%7B%20%5CDelta%20t%7D%20)
<em>where, a and b are the coefficients of de reactant X and product Y, respectively. </em>
(a) Based on the definition above, we can express the rate of reaction (2) as follows:
3O₂(g) → 2O₃(g) (2)
![rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%20%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
(3)
(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:
![rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}](https://tex.z-dn.net/?f=%20rate%20%3D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B1%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5BO_%7B2%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
![+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}](https://tex.z-dn.net/?f=%20%2B%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%20-%5Cfrac%7B2%7D%7B3%7D%20%5Cfrac%7B%5CDelta%5B-1.61%20%5Ccdot%2010%5E%7B-5%7D%5D%7D%7B%20%5CDelta%20t%7D%20)
![\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5CDelta%5BO_%7B3%7D%5D%7D%7B%20%5CDelta%20t%7D%20%3D%201.07%20%5Ccdot%2010%5E%7B-5%7D%20%5Cfrac%7Bmol%7D%7BLs%7D%20)
So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.
Have a nice day!
