The answer is 492.8 g
1. Calculate a number of moles of a sample.
2. Calculate a molar mass of C3H8.
3. Calculate a mass of the sample.
1. Avogadro's number is the number of units (atoms, molecules) in 1 mole of substance: 6.023 × 10²³ units per 1 mole
6.023 × 10²³ atoms : 1 mol =6.72 × 10²⁴ atoms : n
n = 6.72 × 10²⁴ atoms * 1 mol : 6.023 × 10²³ atoms = 1.12 × 10 mol = 11.2 mol
2. Molar mass (Mr) of C3H8 is sum of atomic masses (Ar) of its elements:
Ar(C) = 12 g/mol
Ar(H) = 1 g/mol
Mr(C3H8) = 3 * Ar(C) + 8 * Ar(H) = 3 * 12 + 8 * 1 = 36 + 8 = 44 g/mol
3. Mass (m) of a sample is number of moles (n) multiplied by molar mass (Mr) of C3H8:
m = n * Mr = 11.2 mol * 44 g/mol = 492.8 g
0.5
Explanation:
Given parameters:
Mass of Ca²⁺ = 10g
unknown:
Equivalent weight = ?
Solution:
Equivalent weight that is the amount of electrons which a substance gains or loses per mole.
Ca²⁺ has +3 charge
It lost 2e⁻;
therefore;
In 1 mole of Ca²⁺, we have 2 equivalent weight
1 mol Ca²⁺ = 2eq. wts.
1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt.
Therefore;
10.0 g Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.
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We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
