Calculate the pH of the solution after the addition of the following amounts of 0.0536 M HNO3 to a 60.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

a) 0.00 mL of HNO3; Ph=

b)8.28mL of HNO3; Ph=

c) 42.0ml of HNO3

d) 84.0 mL of HNO3; Ph=

f) 88.5 mL of HNO3; Ph=

A mixture is prepared using 27.00 mL of a 0.0774 M weak base (pKa = 4.635), 27.00 mL of a 0.0581 M weak acid (pKa = 3.189) and 1.00 mL of 0.000116 M HIn and then diluting to 100.0 mL, where HIn is the protonated indicator. The absorbance measured at 550 nm in a 5.000 cm cell was 0.1123. The molar absorptivity (ε) values for HIn and its deprotonated form In– at 550 nm are 2.26 × 104 M–1cm–1 and 1.53 × 104 M–1cm–1, respectively. 1.) What is the pH of the solution? 2.)What are the concentrations of In– and HIn? respectively 3.)What is the pKa for HIn?

Instructions: For each part...Calculate the theoretical yield and percent yield. Then calculate the overall yield.

I am really struggling with this and have been working on it for awhile. Could you please show a step by step for my understanding?

Cyclohexanol: molecular weight 100.16g/mol

Cyclohexanol: density 0.947 g/cm 3

Part A: dehydration of cyclohexanol

Amount of cyclohexanol used (g): 7.4 g

Amount of cyclohexanol obtained (g): 1.611g

Cyclohexene: molecular weight 82.14 g/mol

Cyclohexene: density 0.81 g/cm 3

Part B: Oxidative Cleavage of Cyclohexene

Amount of cyclohexene used (g): 1.34 g

Mass of crude adipic acid (g): .394g

Mass of recrystallized adipic acid (g): 0.036g

Recrystallization percent recovery: 9.14%

Trichloroethene (TCE) is a common groundwater contaminant. In terms of cancer risk, which would be better:

(1) to drink un-chlorinated groundwater groundwater with 10 ppb TCE, or (2) drink surface that, as a result of chlorination, contains 50 ppb chloroform. Assume a 30-year consumption period. The potency factor for TCE is 1.1×10^-2 (mg/kg day)^-1 and the potency factor for chloroform is 6.1×10^-3 (mg/kg day)^-1.

(Risk for TCE = 1.35×10-6)

I know the risk equation is: (Potency Factor)( CDI) but I wasn't sure what volume to use for the calculation since it isnt given

In the distillation apparatus, it is important that the bulb of the thermometer is placed below the sidearm of the still head. Briefly explain why the thermometer placement is so important.

Phosphorus pentachloride decomposes at higher temperatures.

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

An equilibrium mixture at some temperature consists of

3.03 g PCl₅, 208.23 g/mol
4.86 g PCl₃, 137.33 g/mol
3.59 g Cl₂, 70.91 g/mol

in a 1.00-L flask.

If you add 1.31 g of Cl₂, how will the equilibrium be affected and what will the concentration of PCl₅ be when equilibrium is reestablished?

shift left

shift right

no shift will occur

[PCl₅] =   mol/L

When the following skeletal equation is balanced under basic conditions, what are the coefficients of the species shown? SiO32- + Br- --> Br2 + Si Water appears in the balanced equation as a (reactant, product, neither) with a coefficient of . (Enter 0 for neither.) Which species is the reducing agent?

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown? Ag+ + Mg-->Ag + Mg2+ Water appears in the balanced equation as a (reactant, product, neither) with a coefficient of . (Enter 0 for neither.) How many electrons are transferred in this reaction?

A student performed the synthesis of aspirin. After weighing their product, they had 0.501g of product. They took 0.250g of this product and diluted it to 250mL as instructed, and then used 10ml of that stock solution and diluted it to 100mL with 0.025 M Fe(NO₃)_3.Upon analysis with a Spectrophotometer, that solution was found to have an absorbance of .150. Using a Beer's Law plot with the equation A=0.250xC where A is absorbance and C is concentration (1x10^-5M), determine the percent purity of their sample.

I am really struggling with this chemistry problem. I don`t even know how to start..

The problem is about recrystallization:

Two crystalline compounds x and y have the same solubility ratio in water( the solubility is: 1 g/100 mL at 20 degree celsius and 10 g/100 mL at 100 degree celsius) In this example we predict that if the x and y compunds is in the mix, they have the same solubility ratio.

1. From a mix of 10 g x and 1 g y shall clean/pure x be extraxted. How much clean x can we get after a recrystallization?

2. How much clean x can we get after a recrystallization if the mix consist of 10 g x and 8 g y?

Would really appreciate some help:)

for the following compounds:

i) give the correct name and indicate the possible isomers for parts (d, f, g,h,j)

ii) determine the dⁿ configuration, coordination number for the metal and electron count according to the 18 electron rule

iii) calculate the spin only magnetic moment

a) K [Mn(CO)_5]   b) H[AuCl₄]

c) Na[n²-C₂H₄)PtCl₃] d) [(n⁶-C₆H₆)₂Cr]

e) K₂[PdCl₄] f) [Ru(H)(Cl)(CO)(PPh₃)₃]

g) [Co(NO₂)(NH₃)₅]SO₄ h) [CoCl(en)(NH₃)₃]Cl

i) [Ir(phen)(Cl)₄] j) [Cr(en)₃]Cl₃

Construct the titration curve, for the titration of 50.0 mL of 0.125 M trimethylamine with 0.175 M HCl (aq).

A. Calculate the initial pH of the solution in the flask.

B. Calculate the pH after 15 mL of titrant is added.

C. Calculate the pH at the equivalence volume.

D. Calculate the pH after 45 mL of titrant is added.

E. Calculate the pH after 60 mL of titrant is added.

Please show all work

Consider the diprotic acid H₂X, where K_a1 = 1.75 x 10^-5 and K_a2 = 4.3 x 10^-12 at 25 °C. The concentration of HX^- (aq) in a 0.125 M solution of H₂X at 25 °C is approximately

• A. 2.0 x 10^-5 M
• B. 4.5 x 10^-3 M
• C. 1.5 x 10^-3 M
• D. 3.5 x 10^-5 M

• E. none of these

The sulfur in a 5.00 g sample of steel was evolved as H₂S and then collected in a solution of CdCl₂ to produce the precipitate CdS. The CdS was then titrated with excess I₂ and the remaining I₂ was then back titrated with 4.82 ml of 0.0510 M sodium thiosulfate. If the concentration of the I₂ is 0.0600 M and a total of 10.0 ml is added, calculate the %S in the steel.

The relevant reactions are:

CdS + I₂ -> S + Cd²⁺ + 2I^-

I₂ + 2S₂O₃^- -> S₂O₆^2- +2I^-