Answer is: 3,3 mol of <span>nitrous oxide gas is produced in this chemical reaction.
</span>Chemical reaction: N₂ + O₂ → 2NO.
n(N₂) = 1,65 mol.
n(NO) = ?
from reaction n(N₂) : n(NO) = 1 : 2.
1,65 mol : n(NO) = 1 : 2.
n(NO) = 3,3 mol.
n - amount of substance.
Answer: 2.58 days
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant = ?
t = age of sample = 6 days
a = initial amount of the reactant = 1 g
a - x = amount left after decay process
= 0.2 g
a) to find the rate constant
b) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
The half life is 2.58 days
Answer:
a. V = 1000 mL
b. Denisty = 0.022 g/mL
Explanation:
a.
First we need to convert the volume of the Osmium into mL. For that purpose we are given the conversion unit as:
1 mL = 0.1 cL
Hence, the given volume of Osmium will be:
V = Volume of Osmium = 100 cL = (100 cL)(1 mL/0.1 cL) = 1000 mL
<u>V = 1000 mL</u>
b.
The density of Osmium is given by the following formula:
Density = mass/Volume
Denisty = 22 g/1000 mL
<u>Denisty = 0.022 g/mL</u>
Answer:
The MO method for N2+ gives the bond order equal to 2.5. But first, we look at the diagram of molecular orbitals for N2 (the bond order for the nitrogen molecule is 3). the N2+ molecule). That is, the bond order for N2+ is 2.5.
Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
The expression used will be:
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%5D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene = 
= specific heat of liquid benzene = 
= enthalpy change for fusion = 
Now put all the given values in the above expression, we get:
![Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]](https://tex.z-dn.net/?f=Q%3D%5B94.4g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28279-322%29K%5D%2B%5B94.4g%5Ctimes%20-125.6J%2Fg%5D%2B%5B94.4g%5Ctimes%201.51J%2Fg.K%5Ctimes%20%28205-279%29K%5D)
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
