The charge on Pb in Pb(SO3)2 is Lead (IV) Sulfite.
140 g of nitrogen (N₂)
Explanation:
We have the following chemical equation:
N₂ + 3 H₂ -- > 2 NH₃
Now, to find the number of moles of ammonia we use the Avogadro's number:
if 1 mole of ammonia contains 6.022 × 10²³ molecules
then X moles of ammonia contains 6.022 × 10²⁴ molecules
X = (1 × 6.022 × 10²⁴) / 6.022 × 10²³
X = 10 moles of ammonia
Taking in account the chemical reaction we devise the following reasoning:
If 1 mole of nitrogen produces 2 moles of ammonia
then Y moles of nitrogen produces 10 moles of ammonia
Y = (1 × 10) / 2
Y = 5 moles of nitrogen
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of nitrogen (N₂) = 5 × 28 = 140 g
Learn more about:
Avogadro's number
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Answer:
PN₂ = 191.3 Kpa
Explanation:
Given data:
Total pressure of tire = 245.0 Kpa
Partial pressure of PO₂ = 51.3 Kpa
Partial pressure of PCO₂ = 0.10 Kpa
Partial pressure of others = 2.3 Kpa
Partial pressure of PN₂ = ?
Solution:
According to Dalton law of partial pressure,
The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.
Mathematical expression:
P(total) = P₁ + P₂ + P₃+ ............+Pₙ
Now we will solve this problem by using this law.
P(total) = PO₂ + PCO₂ + P(others)+ PN₂
245 Kpa = 51.3 Kpa + 0.10 Kpa + 2.3 Kpa + PN₂
245 Kpa = 53.7 Kpa+ PN₂
PN₂ = 245 Kpa - 53.7 Kpa
PN₂ = 191.3 Kpa
Given :
10 gram sample of propane( C₂H₈ ).
To Find :
The number of moles of propane in given sample.
Solution :
Molecular mass of propane, M = (2 × 12) + ( 1 × 8 ) gram/mol
M = 32 gram/mol
We know, number of moles is given by :
Number of moles, n = m/M
n = 10/32 mol
n = 0.3125 mol
Therefore, number of moles in given sample is 0.3125 mol.
The reaction is second order in AB, so:
![v=k[AB]^2](https://tex.z-dn.net/?f=v%3Dk%5BAB%5D%5E2)
. In the statement, we obtain that
![[AB]=0.104~M](https://tex.z-dn.net/?f=%5BAB%5D%3D0.104~M)
and, at 25 ºC,
. Then:
![v=k[AB]^2\\\\ v=0.0164\cdot0.104^2\\\\ v=0.0164\cdot0.010816\\\\ v\approx0.000177=1.77\times10^{-4}~mol/s](https://tex.z-dn.net/?f=v%3Dk%5BAB%5D%5E2%5C%5C%5C%5C%0Av%3D0.0164%5Ccdot0.104%5E2%5C%5C%5C%5C%0Av%3D0.0164%5Ccdot0.010816%5C%5C%5C%5C%0Av%5Capprox0.000177%3D1.77%5Ctimes10%5E%7B-4%7D~mol%2Fs)
Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:
Since each AB molecule forms one of A and one of B,
. Hence:
.
We'll consider that in the beginning there was not A or B. So,
. Furthermore, since the ratio of AB to A and to B is 1:1,
.
Calculating the time by the expression of velocity:
![v=\dfrac{|\Delta[AB]|}{\Delta t}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_{AB}|}{V}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_A||}{250~mL}\\\\ 1.77\cdot10^{-4}=\dfrac{1}{\Delta t}\cdot\dfrac{0.0026~mol}{0.25~L}\\\\ \Delta t=\dfrac{0.0026}{0.25\cdot1.77\cdot10^{-4}}\\\\ \boxed{\Delta t\approx58.76~s}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B%7C%5CDelta%5BAB%5D%7C%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B%7C%5CDelta%20n_%7BAB%7D%7C%7D%7BV%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B%7C%5CDelta%20n_A%7C%7C%7D%7B250~mL%7D%5C%5C%5C%5C%0A1.77%5Ccdot10%5E%7B-4%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B0.0026~mol%7D%7B0.25~L%7D%5C%5C%5C%5C%0A%5CDelta%20t%3D%5Cdfrac%7B0.0026%7D%7B0.25%5Ccdot1.77%5Ccdot10%5E%7B-4%7D%7D%5C%5C%5C%5C%0A%5Cboxed%7B%5CDelta%20t%5Capprox58.76~s%7D)
