Answer and Explanation:
Iodine have lower atomic mass than tellurium even though the atomic number of iodine is more than the atomic number of tellurium
This is because the atomic weight of any element is the sum of number of proton and number of neutron, even though the number of proton in iodine is more so but the number of neutron is less as compared to tellurium which makes the tellurium of high atomic mass
0.355M x 0.0282L= 0.01 moles of H2SO4. Remember sulphuric acid is diprotic so it will release 2 from each molecule.
<span>So moles of protons = 0.01 x 2 = 0.02 moles of H+ </span>
<span>For neutralization: moles H+ = moles OH- </span>
<span>Therefore moles of NaOH = 0.02 </span>
<span>conc = moles / volume </span>
<span>Conc NaOH = 0.02 / 0.025L = 0.8M </span>
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)

[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:

![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)

=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
<span>Let's </span>assume that the gas has ideal gas behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹)
and T is temperature in Kelvin.<span>
<span>
</span>P = 60 cm Hg = 79993.4 Pa
V = </span>125 mL = 125 x 10⁻⁶ m³
n = ?
<span>
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 25 °C = 298 K
<span>
By substitution,
</span></span>79993.4 Pa<span> x </span>125 x 10⁻⁶ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 298 K<span>
n = 4.0359 x 10</span>⁻³ mol
<span>
Hence, moles of the gas</span> = 4.0359 x 10⁻³ mol<span>
Moles = mass / molar
mass
</span>Mass of the gas = 0.529 g
<span>Molar mass of the gas</span> = mass / number of moles<span>
= </span>0.529 g / 4.0359 x 10⁻³ mol<span>
<span> = </span>131.07 g mol</span>⁻¹<span>
Hence, the molar mass of the given gas is </span>131.07 g mol⁻¹
Hydrogen bonds are approximately 5% of the bond strength of covalent C-C or C-H bonds.
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.