I think it would be C.100.5cm or D.100.5ml hope that helps
Counting atoms in a compound can be done by taking one element at a time and multiplying the subscript of the element and the number of molecules of the compounds. For example, H2O, there are two atoms of H adn 1 atom of oxygen.
CaCl2 and KCl are both salts which dissociate in water
when dissolved. Assuming that the dissolution of the two salts are 100 percent,
the half reactions are:
<span>CaCl2 ---> Ca2+ + 2 Cl-</span>
KCl ---> K+ + Cl-
Therefore the total Cl- ion concentration would be coming
from both salts. First, we calculate the Cl- from each salt by using stoichiometric
ratio:
Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles
Cl / 1 mole CaCl2)
Cl- from CaCl2 = 0.1 moles
Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1
mole KCl)
Cl- from KCl = 0.1 moles
Therefore the final concentration of Cl- in the solution
mixture is:
Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)
Cl- = 0.2 moles / 0.5 moles
<span>Cl- = 0.4 moles (ANSWER)</span>
The empirical formula :
C₁₀H₁₆N₄SO₇
<h3>Further explanation</h3>
Given
6.4 g sample
Required
The empirical formula
Solution
mass C :
= 12/44 x 8.37 g
= 2.28
mass H :
= 2/18 x 2.75 g
= 0.305
mass N = 1.06
mass S :
= 32/64 x 1.23
= 0.615
mass O = 6.4 - (2.28+0.305+1.06+0.615) = 2.14 g
Mol ratio :
= C : H : N : S : O
= 2.28/12 : 0.305/1 : 1.06/14 : 0.615/32 : 2.14/16
= 0.19 : 0.305 : 0.076 : 0.019 : 0.133 divided by 0.019
= 10 : 16 : 4 : 1 : 7
The empirical formula :
C₁₀H₁₆N₄SO₇
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
