1. The solubility of magnesium hydroxide (Mg(OH)2) is approximately 3.70 x10- M atroom temperature.a. Write the aqueous dissolution ionic reaction for magnesium hydroxide. Includestate symbols.b. Calculate the solubility product constant (Ksp) for magnesium hydroxide at roomtemperature. Refer back to the General Formula for “K" presented in the"Equilibrium Lab Manual"c. How many grams of magnesium hydroxide are present in 1.0 L of a saturatedsolution at room temperature?

Answered: Image /qna-images/answer/93d45668-cd37-463e-88d1-17c9958be820.jpg

Answered: 1. The solubility of magnesium…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Consider a pH titration curve at 298 K for 50 mL of 0.10 M hypobromous acid, HBrO (Ka = 2.8 x 10−9), titrated with 0.10 M KOH solution. a. What volume of KOH is needed to reach the half-equivalence point? b. Calculate the pH of the HBrO sample at the half-equivalence point. c. Calculate the equivalence point pH. d. Identify a satisfactory pH indicator for the equivalence point. The pH range of colour change is indicated in parentheses. A) Methyl orange (3.0 - 4.3) B) Bromocresol purple (5.0 - 6.5) C) Alizarin yellow R (10.0 - 12.0) D) Methyl red (4.2 - 6.2)
21 Rx KCI Sımple syrup Cherry syrup aa ad 5 mL Sig: I tsp (15 mEg) tid. x 12 days Note: molecular weight of KCI is 74.5 A How many grams of KCI is required to fill this prescription? D. How much simple syrup in milliliters is required to prepare this prescription? (assume KCI occupies no volume) 9.5
It required 66.66 mL of 2.000 M HBRO3 to titrate (neutralize) 50.00 mL of an Al(OH)3 solution. What is the molarity, M, of the aluminum hydroxide solution ? 3 HBRO3 + Al(OH)3 З Н-0 + Al(BrOз)з Answer a few questions before giving the final answer in the last problem. How many moles of Al(OH)3 are required to neutralize the acid ? A. 0.3999 mol B. 0.1333 mol C. 0.09998 mol D. 0.04444 mol
Calculate the mole ratio of sodium acetate and acetic acid needed to make a buffer of p H 4.68 ( see Table 15.1 for Ka values or Appendices G and H for Ka and Ksp values respectively.
The vapor pressure of component C is 272 torr, and that of component E is 176 torr at 35°C. The C-E solution at 35C with x=0.200 has a vapor pressure of 904.2 torr and vapor composition of x=0.272. 1. Calculate y, for C 2.15 and for E 4.99 : in this solution. 2. Calculate a, for C 2.42 and for E 2.678 in this solution. 3. Calculate p,-p, for component C 2264.38 : (J/mol) and E 857.32 : (J/mol). 4. Calculate AmyG= -1092.22 : (J) for the corresponding ideal solution.
Compare the solubility of lead carbonate in each of the following aqueous solutions: Clear All 0.10 M Pb(CH3C00)2 More soluble than in pure water. 0.10 M (NH4)2CO3 Similar solubility as in pure water. 0.10 Μ KNΟS Less soluble than in pure water. 0.10 Μ ΝΗCH, COO
Considering 200 ml of 0.2 m potassium bromide, answer the following two questions: 64. How many Osmol are present in 1 mol KBr: A) 1. B) 2. 65. How many milliosmolal are presented in the above solution: a. 0.4 d. 80 b. 40 e. 800 mOsm C) 3. c. 0.8
NaOH.2L X.02mol/L = .004 mol, H3PO4.8L X.15 mol/L = .12 mol H₂PO4 HPO4² + H+ .108 mol start A. If 200 mls of 0.02 M NaoH is added to 800 mls of 0.15 M phosphoric acid buffer at pH 8.2, what is the resultant pH? (pK1 = 2.1, Pk2=7.2, pk3=12.2) add .004 mol OH- end .012mol -.004 .008 mol pH= pka + log [A-]/[HA] = 7.2 + log.112/.008 +.004 .112mol pH=8.35 B. If the same amount of NaOH in part A is added to 800 mls of water, what is the resulting pH? .004mol/1L =.004 [OH-] pOH = 2.4, pH -11.6
If we were handed a tuke of 2mg/mL BSA how much is required 20μL of each of the following Concentrations? to make zoul total bring up to volume with dH ₂0 @ 01125 mg/mL (1) 0,150 mg Ⓒ 0.50 mg/mL @ 0.75 mg/mL Ⓒ 1.0mg/ml /ML Can it be solved using C₂ C₁V₁ = C₂ V₂ If CI= V₁ yes, please show work. Vz
Example The reference range for blood pH is 7.35 – 7.42. What is this range expressed as [H+] in nmol l-1? Calculation: pH = - log [H+] 7.35 = - log [H+] [H+] = antilog -7.35 = 4.47 x 10-8 mol.l-1 = 44.7nmol.l-1 similarly: pH 7.42 = 38.0 nmol.l-1 Range [H+] = 38 – 44.7 nmol.l-1 If the blood pH decreases in an acidosis from 7.42 to 7.15, what is the change in [H+] in nmol.l-1? so this is example on the worksheet, but i still understand how to answer the question.
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Can you please calculate the following?

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