Answer choices are:
A.It seldom needs refining because of its detailed accuracy.
B. It clearly depicts the particular set of characteristics being modeled.
C. It is an exact visual representation of the characteristics being modeled.
D. It is the same size and color as the scientific phenomena being demonstrated.
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Correct answer choice is:
B. It clearly depicts the particular set of characteristics being modeled.
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Explanation:
In science, a model is a depiction of an approach, an article or even a method or a scheme that is used to illustrate and interpret happenings that cannot be encountered directly. Models are fundamental to what explorers do, both in their investigation as well as when carrying their information.
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
Answer:
The pH of the 0.100 M solution of hydobromous acid HBrO is 4.843
Explanation:
Here we have the reaction given as follows;
HBrO ⇄ H⁺ + BrO⁻¹
Therefore;
HBrO ⇄ H⁺ + BrO⁻¹
Initial concentration of HBrO = 0.100 M decomposes partly to form x moles of each of H⁺ and BrO⁻¹. That is
HBrO ⇄ H⁺ + BrO⁻¹
Initial 0.1 0 0
Change -x +x +x
Equilibrium 0.1 - x x x
Hence;
![Ka = \frac{[H^+][BrO^{-1}]}{[HBrO]} =\frac{x \times x }{0.1 - x} = \frac{x^2}{0.1 -x} = 2.06 \times 10^{-9}](https://tex.z-dn.net/?f=Ka%20%3D%20%20%5Cfrac%7B%5BH%5E%2B%5D%5BBrO%5E%7B-1%7D%5D%7D%7B%5BHBrO%5D%7D%20%3D%5Cfrac%7Bx%20%5Ctimes%20x%20%7D%7B0.1%20-%20x%7D%20%3D%20%5Cfrac%7Bx%5E2%7D%7B0.1%20-x%7D%20%3D%202.06%20%5Ctimes%2010%5E%7B-9%7D)
(0.1 - x) × 2.06×10⁻⁹ = x²
x² + 2.06×10⁻⁹·x - 0.1 = 0
Factorizing gives;
(x + 1.4354×10⁻⁵)(x - 1.4352×10⁻⁵) = 0
Therefore, x = 1.4352×10⁻⁵ M or -1.4354×10⁻⁵ M
We take the positive value as x is the concentration of the ions in the solution;
From the above [H⁺] = 1.4352×10⁻⁵ M
pH = -log[H⁺] = -log(1.4352×10⁻⁵) = 4.843
The pH of the 0.100 M solution of hydobromous acid HBrO at 298 K where the Ka = 2.06×10⁻⁹ = 4.843.
