Answer:
1. 0.178 moles ; 2. 8x10²³ atoms ; 3. 7.22x10²³ molecules ; 4. 89.6 g ; 5. 1.34x10²² atoms ; 6. 1.67x10²⁵ molecules
Explanation:
1. Mass / Molar mass = Mol
5g / 28 g/m = 0.178 moles
2. 1 molecule of N₂ has 2 atoms, it is a dyatomic molecule.
4x10²³ x2 = 8x10²³ atoms
3. 1 mol of anything, has 6.02x10²³ particles
6.02x10²³ molecules . 1.2 mol = 7.22x10²³
4. 1 atom of C weighs 12 amu.
4.5x10²⁴ weigh ( 4.5x10²⁴ . 12) = 5.24x10²⁵ amu
1 amu = 1.66054x10⁻²⁴g
5.24x10²⁵ amu = (5.24x10²⁵ . 1.66054x10⁻²⁴) = 89.6 g
5. Molar mass NaCl = 58.45 g/m
1.3 g / 58.45 g/m = 0.0222 moles
1 mol has 6.02x10²³ atoms
0.0222 moles → ( 0.0222 . 6.02x10²³) = 1.34x10²²
6. Density of water is 1 g/mL, so 500 mL are contained in 500 g of water
Molar mass H₂O = 18 g/m
500 g / 18 g/m = 27.8 moles
6.02x10²³ molecules . 27.8 moles = 1.67x10²⁵
<span>Data:
pH = 5.2
[H+] = ?
Knowing that: (</span><span>Equation to find the pH of a solution)</span>
![pH = -log[H+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%2B%5D)
<span>
Solving:
</span>
![pH = -log[H+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%2B%5D)
![5.2 = - log [H+]](https://tex.z-dn.net/?f=5.2%20%3D%20-%20log%20%5BH%2B%5D)
Knowing that the exponential is the opposite operation of the logarithm, then we have:
![[H+] = 10^{-5.2}](https://tex.z-dn.net/?f=%5BH%2B%5D%20%3D%2010%5E%7B-5.2%7D)
Answer:
Detail is given below
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.
In A we can see that there is one positive charge and force of attraction is 2.30×10⁻⁸ N and distance is 0.10 nm
In B we can see that negative charge is further away from nucleus because of greater distance thus force of attraction will be less. 0.58×10⁻⁸ N
In C this distance further increases and force also goes in decreasing 0.26×10⁻⁸ N.
Complete Question
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?
Answer:
The pK_a value is
Explanation:
From the question we are told
The volume of base is 
The pH of solution is 
The concentration of the acid is 
From the pH we can see that the titration is between a strong base and a weak acid
Let assume that the the volume of acid is 
Generally the concentration of base

Substituting value


When 13mL of the base is added a buffer is formed
The chemical equation of the reaction is

Now before the reaction the number of mole of base is
![No \ of \ moles[N_B] = C_B * V_B](https://tex.z-dn.net/?f=No%20%5C%20of%20%5C%20moles%5BN_B%5D%20%20%3D%20%20C_B%20%2A%20V_B)
Substituting value

Now before the reaction the number of mole of acid is

Substituting value


Now after the reaction the number of moles of base is zero i.e has been used up
this mathematically represented as

The number of moles of acid is


The pH of this reaction can be mathematically represented as
![pH = pK_a + log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%20%20%3D%20pK_a%20%2B%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
Substituting values
