Answer:- 21.4 grams of 
are formed.
Solution:- The balanced equation is:
From this equation, lithium and nitrogen reacts in 6:1 mol ratio. Limiting reactant gives the theoretical yield of the product. We will calculate the grams of the product for the given grams of both the reactants and see which one of them gives the limited amount of the product. This limited amount of the product will be the theoretical yield.
The molar mass of Li is 6.94 gram per mol and for 
It is 28.02 gram per mol. The molar mass of is 34.83 gram per mol. The calculations for the grams of the product for given grams of both the reactants are shown below:
= 
= 
From above calculations, Li gives least amount of the product. So, 21.4 g of are formed.
Answer:
Explanation:
Hello there!
In this case, given the solubilization of cadmium (II) hydroxide:
The solubility product can be set up as follows:
![Ksp=[Cd^{2+}][OH^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCd%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:
Regards!
STP condition mean we have P=1 atm. T=273K. R=ideal gas constant, but make sure you use the one that has the same units of pressure, temperature that you are using. In this case R=0.0821 L*atm K^-1mol^1. You are provided with n=2.1 moles.
V=nRTP
Input your values and solve.
Answer:
2
3
Explanation:
To infer the last energy of the given atoms, we need to write their electronic configuration:
For N = 1S² 2S² 2P³
Mg = 1S² 2S² 2P⁶ 3S²
The energy levels are usually designated as;
n = 1
n = 2
n = 3
n =4
For N, the last energy level is 2
Mg, the last energy level is 3
We can also determine this number by the periods the atoms can be found.
<h2>
→
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Explanation:
Ethanol can be oxidized to ethanal or acetaldehyde which is further oxidized to acid that is acetic acid.
→ [oxidation by loss of hydrogen]
-
An oxidizing agent potassium dichromate(VI) solution is used to remove the hydrogen from the ethanol.
- An oxidizing agent used along with dilute sulphuric acid for acidification.
Acetaldehyde can also be reduced back to ethanol again by adding hydrogen to it by using a reducing agent that is sodium tetrahydro borate, NaBH4.
- The oxidation of aldehydes to carboxylic acids can be done by the two-step process.
- In the first step, one molecule of water is added in the presence of a catalyst that is acidic.
- There is a generation of a hydrate. (geminal 1,1-diol).
→ 
[reduction by the gain of electrons]
Here, the oxidizing agent used is
in the presence of acetone.
