Answer:
9.1 mg/dL
Explanation:
A normal distribution is symmetrical, the area below the curve of negative Z-scores is equal to the area below the curve of positive Z-scores, Z measures the statistic observed and the parameter of the population.
And it is:
Z = (x - μ)/σ
where μ is the mean (9.5 mg/dL), x is the borderline, and σ is the standard deviation, which is 0.3 mg/dL.
The Z-score can be calculated in excel by the function =NORMSINV, and the factor will be the percente considered (10% = 0.1)
NORMSINV(0.1) = -1.28155
So,
-1.28155 = (x - 9.5)/0.3
x - 9.5 = -0.384465
x = 9.1 mg/dL
<span>The products of the light-dependent reactions are used to help 'fuel' the light-independent reactions.
</span><span>Example:
NADPH and ATP are produced during the light-dependent reaction for use in the light-independent reaction (the Calvin Cycle). </span>
Atomic mass is the answer!!
Explanation: An elements atomic number won’t be able to change
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6