Answer:
2.7 × 10⁻⁴ bar
Explanation:
Let's consider the following reaction at equilibrium.
SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.
Kp = pSbCl₃ × pCl₂ / pSbCl₅
pCl₂ = Kp × pSbCl₅ / pSbCl₃
pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22
pCl₂ = 2.7 × 10⁻⁴ bar
Answer: Adenine and guanine are the two purines and cytosine, thymine and uracil are the three pyrimidines. The main difference between purines and pyrimidines is that purines contain a sixmembered nitrogencontaining ring fused to an imidazole ring whereas pyrimidines contain only a sixmembered nitrogencontaining ring. They both are types or categories of nitrogen containing bases present in nuclei acids of DNA and RNA.
Purines are 2 Ring or Carbon Ring, Nitrogen containing bases. That consist of these 2 rings next placed next to each other. These examples include - Adenine and Guanine.
Pyrimidines are 1 or single Ring Nitrogen containing structures. There are 3 nitrogenous bases that are categorized as pyrimidines. Cytosine, Thymine, and Uracil.
Answer:Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.
Explanation:
Answer:
To prepare a 1 M solution, slowly add 1 formula weight of compound to a clean 1-L volumetric flask half filled with distilled or deionized water. Allow the compound to dissolve completely, swirling the flask gently if necessary.
Explanation:
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Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
