1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.
2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.
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
Answer: 37 protons, 48 neutrons, and 36 electrons.
Explanation: I assume "Revision" is Rubidium. It is atomic number 37, so it has 37 protons. It's atomic mass is 85, so it has an average of (85-37) = 48 neutrons. It has a charge of +1, so one electron of the original 37 has left.
Answer:
2 chlorine atoms.
Explanation:
H₂ + Cl₂ → 2HCl
The Law of Conservation of Mass states that matter can neither be created nor destroyed in a chemical reaction.
If 2 molecules of Hydrogen and 2 molecules of Chlorine is present in the reactants, the same no. of atoms should also be present in the product.
∴ the no. of chlorine atoms in the product is 2.
Answer:
you will go no where as far as you wish because there is no air resistance or friction, causing you to not stop. In the real world, you would stop a little bit after you got done with the hill because you have friction and air Resistance to bring you to a stop. There is a huge difference because in real life, there is a lot of friction and air resistance in the real world, which will eventually bring someone to a stop no matter how fast they are going.
