<span>The half-life of a first-order reaction is determined as follows:
</span>t½<span>=ln2/k
From the equation, we can calculate the </span><span>first-order rate constant:
</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³
When we know the value of k we can then calculate concentration with the equation:
A₀ = 2 g/100 mL
t = 2.5 h = 150min
A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml
= 6.3 × 10⁻⁴ mg / 100ml
Answer:
Dalton's atomic model and Rutherford model
Explanation:
There were differences in the models:
Thompson's experiment showed that atoms contained tiny negatively charged particles called electrons.
On the other hand, Rutherford gold experiment led to the conclusion that the atom is a empty space with tiny, dense and positively charged nucleus.
Dalton thought that atoms were the smallest units of matter that could not be broken down further. This assumption continued to hold until it was later discovered that the atom was composed of subatomic particles.
Answer:
0.12 M hydrofluoric acid + 0.17 M potassium fluoride
Explanation:
To make a buffer, you must to have an aqueous mixture of a weak acid and its conjugate base or vice versa.
Knowing that:
0.32 M calcium chloride + 0.27 M sodium chloride: <em>is not a good buffer system </em>because CaCl₂ and NaCl are both salts.
0.35 M ammonia + 0.36 M calcium hydroxide <em>is not a good buffer system </em>because ammonia is a weak base but calcium hydroxide is a strong base
0.19 M barium hydroxide + 0.28 M barium chloride <em>is not a good buffer system </em>because Ba(OH)₂ is a strong base.
0.12 M hydrofluoric acid + 0.17 M potassium fluoride <em>is a good buffer system </em>because HF is a weak acid and KF (F⁻ in aqueous medium), is its conjugate base
0.20 M hydrobromic acid + 0.22 M sodium bromide <em>is not a good buffer system </em>because HBr is a strong acid.
Answer:
The correct answer to the question is
D. 3 miles (4.5 kilometers)
Explanation:
The Carbonate compensation depth or (CCD) is the ocean depth at which calcite, (calcium carbonate) dissolves. At the CCD, the solvation rate of calcite is greater than the supply rate, such that all calcite are consumed.
The carbonate compensation depth varies in different parts of the ocean and can be reached at about 3 miles or 4.5 Kilometers.
Answer:
D
Explanation:
D. V1P1 / T1=V2P2 / T2 is correct
