Questions
Combustion analysis of 17.471 g of trioxane produces 10.477 g of water and 25.612 g of...

Combustion analysis of 17.471 g of trioxane produces 10.477 g of water and 25.612 g of carbon dioxide. The molar mass of trioxane is approximately 90. g/mol. Determine the molecular formula of this compound.

In: Chemistry

The choice of an appropriate solvent is crucial for a successful recrystillization. Describe how you will...

The choice of an appropriate solvent is crucial for a successful recrystillization. Describe how you will determine the proper recrystillization solvent for the Aldol product.

In: Chemistry

Basic understanding how double bonds absorb light. Overall chemical reaction of ‘fixing’ carbon dioxide.

Basic understanding how double bonds absorb light. Overall chemical reaction of ‘fixing’ carbon dioxide.

In: Chemistry

In the lab, 1. we put sodium oxalate into lead nitrate 2. we put sulphuric acid...

In the lab,

1. we put sodium oxalate into lead nitrate

2. we put sulphuric acid into lead nitrate

After both solution has precipated, we removed the liquids and pour nitric acid to each of precipitates in seperate test tubes.

The observation was that

First tube was much thinner and the second tube become much thicker (in terms of white-ish color) after adding nitric acid.

What is the reason for this difference?

In: Chemistry

I posted this exact question a few days ago and some doofus said it wasn't clear....

I posted this exact question a few days ago and some doofus said it wasn't clear. I don't know how to make it any more clear. This is all the information the question gives. There are 5 parts, A-E. Please answer them all. Thank you.

Consider the titration of a 20.0 −mL sample of 0.110 M HC2H3O2 with 0.125 M NaOH. Determine each quantity:

Part A)

The initial pH = ?

Part B)

The volume of added base required to reach the equivalence point.

V = ? mL

Part C)

The pH at 4.00 mL of added base.

pH = ?

Part D)

the pH at one-half of the equivalence point.

pH = ?

Part E)

The pH at the equivalence point.

pH = ?

In: Chemistry

Consider an ideal monatomic gas og N indistinguishable particles, mass m each and total energy U,...

Consider an ideal monatomic gas og N indistinguishable particles, mass m each and total energy U, confined to a one dimnensional channel of length L.

Calculate the multiplicity of the 1-D gas and its entropy, following the line of reasoning used in deriving these quantites in the 3-D case.

In: Chemistry

Part A Calculate Kc for the reaction below. I2(g)⇌2I(g)Kp=6.26×10−22 (at 298 K) Express your answer using...

Part A

Calculate Kc for the reaction below.
I2(g)⇌2I(g)Kp=6.26×10−22 (at 298 K)

Express your answer using three significant figures.

Kc=?

Part B

Calculate Kc for the reaction below.
CH4(g)+H2O(g)⇌CO(g)+3H2(g)Kp=7.7×1024 (at 298 K)

Express your answer using two significant figures.

Kc=

In: Chemistry

Consider the titration of a 28.0 −mL sample of 0.175 M CH3NH2 with 0.155 M HBr....

Consider the titration of a 28.0 −mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following. 1. the inital pH 2. the volume of added acid required to reach the equivalence point 3. the pH at 6.0 mL of added acid 4. the pH at one-half of the equivalence point 5. the pH at the equivalence point 6. the pH after adding 5.0 mL of acid beyond the equivalence point

In: Chemistry

You prepare a 0.200M solution of a monoprotic buffer at pH 7.22. The pKa of this...

You prepare a 0.200M solution of a monoprotic buffer at pH 7.22. The pKa of this buffer is 7.53 at room temperature.

a. What is the molarity of the acid and conjugate base necessary to make the buffer. [HA]=? [A-]=?

b. A separate 1.0 M stock solution of the acid and conjugate base is made/ How many mL of each will you use to prepare 1.0 L of the buffer in part a? mL of HA=? mL of A-=?

c. You add 29mL of 2.3 M NaOH to your 1.0L of buffer. Is the buffer capacity able to adequately handle this new addition of NaOH? What will be your new pH?

In: Chemistry

Compound A is a D-aldopentose that can be oxidized to an optically active aldaric acid B....

Compound A is a D-aldopentose that can be oxidized to an optically active aldaric acid B. On Kiliani-Fischer chain extension, A is converted into C and D; C can be oxidized to an optically inactive aldaric acid E, but D is oxidized to an optically active aldaric acid F. What is the structure of compound E?

In: Chemistry

When 22.63 mL of aqueous NaOH was added to 1.136 g of cyclohexylaminoethanesulfonic acid (FM 207.29,...

When 22.63 mL of aqueous NaOH was added to 1.136 g of cyclohexylaminoethanesulfonic acid (FM 207.29, pKa = 9.39, structure in the table) dissolved in 41.37 mL of water, the pH was 9.24. Calculate the molarity of the NaOH.

In: Chemistry

Draw a simple molecular orbital diagram for the following molecules. O22.... N2+.... C22- '''a. Is the...

Draw a simple molecular orbital diagram for the following molecules. O22.... N2+.... C22- '''a. Is the ion paramagnetic or diamagnetic? b. What is the bond order in the ion? c. Would these ions have longer, shorter, or the same bond length as their neutral counterpart? d. Would these ions have stronger, weaker, or equal bonding strength as their neutral counterparts

In: Chemistry

1) Consider the reaction: A (aq) ⇌ B (aq) at 253 K where the initial concentration...

1) Consider the reaction:


A (aq) ⇌ B (aq)

at 253 K where the initial concentration of A = 1.00 M and the initial concentration of B = 0.000 M. At equilibrium it is found that the concentration of B = 0.523 M. What is the maximum amount of work that can be done by this system when the concentration of A = 0.859 M?

2) Consider the following reaction at 257 K:

2 A (aq) + 1 B (aq) → 2 C (aq) + 2 D (aq)

An experiment was performed with the following intitial concentrations: [A]i = 1.37 M, [B]i = 2.19 M, [C]i = 0.47 M, [D]i = 0.31 M. The reaction was allowed to proceed until equilibrium was reached at which time it was determined that [A] = 0.57 M. What is ΔGnonstandard at the initial conditions? (I got -14.16kJ but it said i was wrong)

3) Consider the cell described below at 287 K:

Sn | Sn2+ (1.27 M) || Pb2+ (2.49 M) | Pb

Given EoPb2+ + 2 e-→Pb = -0.131 V and EoSn2+ + 2 e-→Sn = -0.143 V. What will the concentration of the Pb2+ solution be when the cell is dead?

4)A concentration cell is built based on the reaction:

2 H+ + 2 e- → H2

The pH in one of the half cells is -0.03743, while the pH in the other is 2.804. If the temperature of the overall cell is 296 K, what is the potential? (Faraday's constant is 96,485 C/mol e-)

In: Chemistry

When creating a titration curve for a weak base, the pH of the initial solution requires...

When creating a titration curve for a weak base, the pH of the initial solution requires setting up a table showing the initial, change, and equilibrium values for each species and plugging these into the equilibrium constant expression. The expression for the bicarbonate ion is Ka1=[H2CO3][H+][HCO3−] For an anion that can both hydrolyze and produce H+, the pH of a concentrated solution can be more easily approximated using the equation pH=12(pKa1+pKa2) During the titration before the equivalence point, provided that the concentration of acid is significantly more than the concentration of base, the Henderson-Hasselbalch equation can be used to approximate the pH: pH=pKa+log[base][acid] At the equivalence point, the solution is no longer a buffer, but contains the weak acid H2CO3. The change in concentration of the bicarbonate ion is significant, and the equilibrium constant expression for Ka1 must again be used to find the concentration of hydronium ions. After the equivalence point, the strong acid will control the pH.

Part A

A 10.0-mL sample of 1.0 M NaHCO3 is titrated with 1.0 M HCl (hydrochloric acid). Approximate the titration curve by plotting the following points: pH after 0 mL HCl added, pH after 1.0 mL HCl added, pH after 9.5 mL HCl added, pH after 10.0 mL HCl added (equivalence point), pH after 10.5 mL HCl added, and pH after 12.0 mL HCl added.

In: Chemistry

Part A Calculate the change in pH that results from adding 0.190 mol NaNO2 to 1.00...

Part A

Calculate the change in pH that results from adding 0.190 mol NaNO2 to 1.00 L of 0.190 M HNO2(aq). (Ka(HNO2)=7.2⋅10−4) Express your answer using two decimal places.

Part B

Calculate the change in pH that results from adding 0.190 NaNO3 to 1.00 L of 0.190 HNO3(aq). Express your answer using two decimal places.

In: Chemistry