Answer:
Chemistry plays an important and useful role towards the development and growth of a number of industries. This includes industries like glass, cement, paper, textile, leather, dye etc. We also see huge applications of chemistry in industries like paints, pigments, petroleum, sugar, plastics, Pharmaceuticals.
Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
Answer:the pH is 12
Explanation:
First We need to understand the structure of trimethylamine
Due to the grades of the bond in the nitrogen with a hybridization sp3 is 108° approximately, then is generated a dipole magnetic at the upper side of the nitrogen, this dipole magnetic going to attract a hydrogen molecule of the water making the water more alkaline
C3H9N+ H2O --> C3H9NH + OH-
![k=\frac{[C3H9NH]*[OH-]}{[C3H9N]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B%5BC3H9NH%5D%2A%5BOH-%5D%7D%7B%5BC3H9N%5D%7D)
Then:
The concentration of the trimethylamine is 0.3 and the concentration of the ion C3H9NH is equal to the OH- relying on the stoichiometric equation. We could find the concentration of the OH- ion with the square root of the multiplication between k and the concentration of trimethylamine
[OH-]=
[OH-]=0.01
pH=14-(-log[OH-])
pH=12
Answer:
Nitrogen
Explanation:
Nitrogen has 5 Valence Electrons.
And its atomic mass is also less than Bismuth.
There are other elements in the 5th row but this one has the lowest atomic mass compared to the others.
1.70 × 10³ seconds
<h3>Explanation </h3>
+ 2 e⁻ → 
It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.
Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains 
of Co atoms. It would take 
of electrons to reduce cobalt (II) ions and produce the 
of cobalt atoms.
Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of 
. A current of 0.961 A delivers 0.961 C of charge in one single second. It will take 
to transfer all these charge and deposit 0.500 g of Co.
