Answer:
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
2Fe + 3Cl2 → 2FeCl3
Explanation:
1. (SO4) 3 you see this 3 it means that 3 must be behind H2SO4. So now it's 3H2SO4.
2. If 3 is now behind one H2, it must be behind the other.
So now it's 3H2.
3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.
1. FeCl3 has 3 ahead of Cl, and Cl2 has 2. Which means that behind FeCl3 goes 2, and behind Cl2 goes 3 so now we have equated all Cl.
2. Since it is now 2FeCl3, we know that there must be 2 in the second Fe. It's 2Fe now.
Answer: R & D is the correct option.
Explanation:
Research and development department of a chemical industry requires individuals holding advanced degrees because :
- This will help the chemical industry to come up with new developments and advancements in preparation of the chemical products like :green methods to prepare certain chemicals which usually involves handling of toxic substances.
- Formulation and development of the new chemical substitutes
- Improvement of chemical product at microscopic level.
10 moles of H2 reacts with 5 moles of O2 creating 10 moles of water.
Water has the molar mass 18 g/mole
10 · 18g = 180 g
180 g of water can be produced from 10 moles of hydrogen gas.
Answer:
Amount of pyridine required = 0.0316 M
Explanation:
pH of a buffer solution is calculated by using Henderson - Hasselbalch equation.
![pH=pK_a+log\frac{[Conjugate\ base]}{[weak\ acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BConjugate%5C%20base%5D%7D%7B%5Bweak%5C%20acid%5D%7D)
Pyridinium is a weak acid and in the presence of its conjugate base, it acts as buffer.
Henderson - Hasselbalch equation for pyridine/pyridinium buffer is as follows:
![pH=pK_a+log\frac{[Py]}{PyH^+]}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BPy%5D%7D%7BPyH%5E%2B%5D%7D)
pH = 4.7
(Pyridinium)=0.100 M
Substitute the values in the formula
![pH=pK_a+log\frac{[Py]}{PyH^+]}\\4.7=5.2 log\frac{[Py]}{0.100}](https://tex.z-dn.net/?f=pH%3DpK_a%2Blog%5Cfrac%7B%5BPy%5D%7D%7BPyH%5E%2B%5D%7D%5C%5C4.7%3D5.2%20log%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D)
![4.7-5.2=log\frac{[Py]}{0.100} \\-0.5=log\frac{[Py]}{0.100}\\\frac{[Py]}{0.100}=antilog -0.5\\\frac{[Py]}{0.100}=0.316](https://tex.z-dn.net/?f=4.7-5.2%3Dlog%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%20%5C%5C-0.5%3Dlog%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%5C%5C%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%3Dantilog%20-0.5%5C%5C%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%3D0.316)
![\frac{[Py]}{0.100} =0.316](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPy%5D%7D%7B0.100%7D%20%3D0.316)
[Py]=0.0316\ M
Amount of pyridine required = 0.0316 M
