The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
Answer:
D. Intramolecular covalent bond
Explanation:
Compound D is structurally more rigid as a result of intramolecular covalent bonding. The forces that hold together atoms within a compound are greater as compared to forces holding two molecules together (intermolecular bonding). On the other hand Hydrogen bonds are weaker as compared to covalent bonds. Covalent bonds involve the sharing of electrons between two atoms and Hydrogen bonds are formed between a highly electronegative atom like oxygen, Flourine,Chlorine to hydrogen.
<span>440 g
First, determine the volume of each sheet. And it's easier if each dimension is using the same unit of measure. So each sheet is 28 cm by 22 cm by 0.30 cm. Multiply them together
28 cm * 22 cm * 0.30 cm= 184.8 cm^3
Since we have 2 identical sheets, double the total volume
184.8 cm^3 * 2 = 369.6 cm^3
Now multiple the volume by the density
369.6 cm^3 * 1.2 g/cm^3 = 443.52 g
Round the results to 2 significant digits since all of the given figures are only 2 significant digits long.
443.52 g = 440 g</span>
Answer : The pH of 0.289 M solution of lithium acetate at 
is 9.1
Explanation :
First we have to calculate the value of 
.
As we know that,
where,
= dissociation constant of an acid = 
= dissociation constant of a base = ?
= dissociation constant of water = 
Now put all the given values in the above expression, we get the dissociation constant of a base.
Now we have to calculate the concentration of hydroxide ion.
Formula used :
![[OH^-]=(K_b\times C)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%28K_b%5Ctimes%20C%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
where,
C is the concentration of solution.
Now put all the given values in this formula, we get:
![[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%285.5%5Ctimes%2010%5E%7B-10%7D%5Ctimes%200.289%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![[OH^-]=1.3\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.3%5Ctimes%2010%5E%7B-5%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
Now we have to calculate the pH.
Therefore, the pH of 0.289 M solution of lithium acetate at is 9.1
A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.
