Answer:
0.00125 moles H₃X
Solution and Explanation:
In this question we are required to calculate the number of moles of triprotic acid neutralized in the titration.
Volume of NaOH used = final burette reading - initial burette reading
= 39.18 ml - 3.19 ml
= 35.99 ml or 0.03599 L
Step 1: Moles of NaOH used
Number of moles = Molarity × Volume
Molarity of NaOH = 0.1041 M
Moles of NaOH = 0.1041 M × 0.03599 L
= 0.00375 mole
Step 2: Balanced equation for the reaction between triprotic acid and NaOH
The balanced equation is;
H₃X(aq) + 3NaOH(aq) → Na₃X(aq) + 3H₂O(l)
Step 3: Moles of the triprotic acid (H₃X used
From the balanced equation;
1 mole of the triprotic acid reacts with 3 moles of NaOH
Therefore; the mole ratio of H₃X to NaOH is 1 : 3.
Therefore;
Moles of Triprotic acid = 0.00375 mole ÷ 3
= 0.00125 moles
Hence, moles of triprotic acid neutralized during the titration is 0.00125 moles.
Answer:
The structure is shown below.
Explanation:
To draw a structure first we need to know its molecular formula, which is C2H6SO for dimethyl sulfoxide. The central atom is sulfur, which is bonded to an oxygen and with two methyl groups (CH3).
Sulfur has 6 electrons in its valence shell, as so oxygen. To complete the octet of oxygen, 2 electrons will be shared by sulfur with it. So, it remains 4 electrons at the central atom. Carbon has 4 electrons in its valence shell, so it needs more 4 to be stable, and is already sharing 3 electrons with the hydrogens, thus, sulfur will share one electron with each one of them.
So, it will remain 2 nonbonding electrons in the central atom. According to the VSPER theory, to minimize formal charges, the structure would be a trigonal pyramid, but, the double bonding with oxygen has a large volume, then the geometry will be trigonal, as shown below.
Answer:
Water acts as a base in the presence of a strong acid
Explanation:
Water,being an amphoteric compound, can act both as an acid and as a base.
In the presence of an acid , water acts as a base but in the presence of a base, water acts as an acid.
<span>We can solve this problem by assuming that the decay of
cyclopropane follows a 1st order rate of reaction. So that the
equation for decay follows the expression:</span>
A = Ao e^(- k t)
Where,
A = amount remaining at
time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560
M </span><span>
<span>k = rate constant
t = time = 1.50 hours or 5400 s </span></span>
The rate constant should
be given in the problem which I think you forgot to include. For the sake of
calculation, I will assume a rate constant which I found in other sources:
k = 5.29× 10^–4 s–1 (plug in the correct k value)
<span>Plugging in the values
in the 1st equation:</span>
A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )
A = 3.218 <span>× 10^–4 M (simplify
as necessary)</span>
