on a hot summer day, water droplets often form on the outside of a cold glass. the water droplets form when water vapor in the a
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Answer:
The absolute uncertainty is approximately 1.69 × 10⁻³
Explanation:
The volume needed for NaOH needed to make the solution = 16.66 ml
The wt% of the added NaOH = 53.4 wt%
The volume of the NaOH to be prepared = 2.00 L
The concentration of the NaOH to be prepared = 0.169 M
The molar mass of NaOH = 39.997 g/mol
Therefore, 100 g of sample contains 53.4 g of NaOH
The mass of the sample = 16.66 × 1.52 = 25.3232 g
The mass of NaOH in the sample = 0.534 × 25.3232 = 13.5225888 g ≈ 13.52 g
Therefore;
The number of moles of NaOH = 13.52/39.9971 = 0.3381 moles
Therefore, we have 0.3381 moles in 2.00L solution, which gives;
The number of moles per liter = 0.3881/2 = 0.169045 moles/liter
The molarity ≈ 0.169 M
The absolute uncertainty, u(c) is given as follows;
The absolute uncertainty, u(c) ≈ 1.69 × 10⁻³.
At equilibrium the concentrations of:
[HSO₄⁻] = 0.10 M;
[SO₄²⁻] = 0.037 M;
[H⁺] = 0.037 M;
There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.
HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.
R ⇄
I
C
E
×
for
. As a result,
is large. It is no longer valid to approximate that
at equilibrium is the same as its initial value.
×
×
Solving the quadratic equation for since
represents a concentration;
Then, round the results to 2 significant figure;
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