Sunday

April 19, 2015

April 19, 2015

Total # Posts: 127

**chemistry who helps me step by step**

25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. calculate the percentage of fe204 7h20 in the sample. My calculation 0.02x...
*January 11, 2014*

**chemitry who helps me **

you are given a solid that is a mixture of na2s04 and k2s04. A0.215g sample of the mixture is dissolved in water. An excess of an aqueous solution of bacl2 is added. The bas04 that is formed is filtered dried and weight. Its mass is 0.299g what mass of s04 ions is the sample ...
*January 5, 2014*

**chemistry Damon check my calculation**

25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of this solution required 20ml of n/10 kmn04 solution for complete oxidation. Calculate the percentage of f2s047h20 in ...
*January 4, 2014*

**math who helps me step by step**

A mixture of pure k2cr207 and pure kmn04 weighing 0.561g was treated with excess of ki in acidic medium. Iodine liberated required 100ml of 0.15M of sodium thiosulphate solution for exact oxidation. The right answer 43.67 k2cr207 and 56.33 kmn04 My calculation 0.1L x 0.15M = 0...
*January 4, 2014*

**math who helps me step by step**

25g of a sample of ferrous sulphate was dissolved in water contains dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculte the percentage of f2s04 7 h20 in the sample My calculation 0.02 x 0...
*January 4, 2014*

**Damon check step by step**

25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculate the percentage of f2s047h20 in the sample The right answer 88.96...
*January 2, 2014*

**math (Damon) check step by step for me**

A mixture of pure k2cr207 and pure kmno4 weighing 0.561g was treated with excess of ki. In acidic medium. Iodine liberated required 100ml of 0.15M of sodium thiosulphate solution for exact oxidation. The right answer 43.67 k2cr207 and 56.33 kmn04 K2cr207 = 294 (X) kmno4 = 158...
*January 2, 2014*

**math (Damon) clearly step by step for me **

3.2g of a mixture of kn03 and nan03 was heated to constant weight which was found to be 2.64g. What is the % kn03 in mixture? Who helps me clearly step by step for me. Thank you .
*January 1, 2014*

**math who me step by step**

25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculate the percentage of fes047h20 in the sample the right answer 88.96...
*January 1, 2014*

**math who helps me step by step **

A mixture of pure k2cr207 and pure kmn04 weighing 0.561g was treated with excess of ki in acidic medium. iodine liberated required 100ml of 0.15M of sodium thiosulphate solution for exact oxidation. The right answer k2cr207 43.67% kmn04 56.33% K2cr207 =294 kmn04 = 158 x/294 + ...
*January 1, 2014*

**math steve for your help**

25g of a sample of ferrous sulphate was dissolved in water containing dilute h2s04 and the volume made up to one litre. 25ml of this solution required 20ml of N/10 kmn04 solution for complete oxidation. Calculate the percentage of fes047h20 in the sample The answer 88.96% My ...
*December 29, 2013*

**math Steve for your help**

3.2g of a mixture of kn03 and nan03 was heated to constant weight which was found to be 2.64g. What is the % kn03 in mixture the answer 44.22%
*December 29, 2013*

**math Steve form your help**

50gm of a mixture of ca(OH)2 is dissolved in 50ml of 0.5N hcl solution, the excess of hcl was titrated with 0.3N -naoh. The volume of naoh used was 20cc. Calculate % purity of ca(OH)2 The answer 41.35% Who helps me step by step
*December 29, 2013*

**steve for your solution**

you need to use 3 solutions of drug H containing respectively 50%, 20% and 5% to make 600ml of 10% drug H. How much of the 50% will you use?
*December 28, 2013*

**math Steve for your solution**

0.224g sample that contained only bacl2 and kbr required 19.7ml of 0.1M ango3 to reach the end point. Calculate the percentage of each compound present in the sample. The right answer bacl2 32.4% KBR 67.7% You need to use 3 solution of drug H containing respectively 50%, 20% ...
*December 28, 2013*

**math (steve for your help)**

0.224g sample the contained only bacl2 and kbr required 19.7ml of 0.1M agn03 to reach the end point. Calculate the percent of each compound present in the sample. Steve want you step by step for me.
*December 27, 2013*

**math Steve helps me**

1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8 m Hcl. The excess of acid required 16ml of 0.25< NAOH for neutralization. Calculate the percentage of cac03 and mgc03 in the sample The right answer mgc03 52.02% and caco 47.98% You need to use 3 solution of...
*December 25, 2013*

**Math**

1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgc03 in the sample Steve helps me step by step
*December 25, 2013*

**math**

Determine the volume of diluted nitric acid D1.11g/ml, 19% hn03. That can be prepared by diluting with water 50ml of conc hn03 d= 1.42g/ml, 69.8% I do not still understand how 183.68ml I want you clear explain step by step for me.
*December 25, 2013*

**math**

Determine this volume of diluted nitric acid d = 1.11g/ml, 19% hn03. That can be prepared by diluting with water 50ml of conc hn03 d=1.42g/ml, 69.8% I do not still understand how to get 19% is .2722 , the right answer is 183.68ml
*December 24, 2013*

**math**

Determine the volume of diluted nitric acid d= 1.11g/ml, 19% hn03. That can be prepared by diluting with water 50ml of conc. hn03 d=1.42g/ml, 69.8% The answer 183.68ml who help step by step
*December 24, 2013*

**math**

50gm of a sample of ca(oh)s is dissolved in 50ml of 0.5N hcl solution. The excess of hcl was titrated with 0.3N -naoh. The volume of naoh used was 20cc. Calculate of naoh used was 20cc. Calculate % purity ca(oh)2 Who help me step by step
*December 22, 2013*

**math**

0.224g sample that contained only bacl2 and kbr required 19.7ml of 0.1M agn03 to react the end point. Calculate the percent of each compound present in the sample
*December 21, 2013*

**math**

calculate the number of molecule h2s04 present in 250ml of 98% by mass of sulphuric solution. Density of solution is 1.84g/ml Answer 4.6N How do calculation step by step
*December 21, 2013*

**math**

1.64g of a mixture of cac03 and mhco3 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgc03 in the sample Who helps me
*December 21, 2013*

**math**

1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgc03 in the sample The answer 52.02% mgc03 caco 47.98% Bob helps me step by step Thank you
*December 8, 2013*

**math**

1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgco3. The right answer mgc03 52.02% caco 47.98% Who help for me step by step Thank you
*December 7, 2013*

**maths**

A person drives 500 km in 6 months. How many km will that person drive in 3 years?
*November 19, 2013*

**maths**

A community college has 2,000 students and 50 instructors. Next year the enrollment will be 6,000. How many new instructors should be hired if the college wants to keep the same student to instructor ratio?
*November 19, 2013*

**chemistry (Pls Bob check for me)**

what mass of 25% ba(n03)2 solution contain 40 grams of ba(n03)2 ? Answer 160g My calculation 40g/261.37g/mol = 0.15305/0.25g = 0.612 This answer is not right Who helps
*April 28, 2013*

**chemistry (Pls Bob check for me)**

Aspirin c9h8oh molar mass = 180.1g/mol can be prepared by the reaction of salicylic acid c7h603 molar mass = 13801 g/mol with acetic anhydride c6h603, according to the following equation. A chemist starts with 2 gram of salicylic acid reacts it with excess acetic anhydrate, ...
*April 28, 2013*

**math**

Give 0.7ml of metoclopramide syrup very six hours anwer 1 tsp oral syringe with 1 ml marking Who help
*April 21, 2013*

**pharmacy math**

which would be the most appropriate measuring device for the following prescription: Give 0.7ml of metoclopramide syrup every six hours answer 1 tsp oral syringe with 1 ml marking I am not sure answer Bob help
*April 14, 2013*

**chemistry (Pls Bob check for me)**

0.5g of fuming h2s04 oleum is diluted with water. This solution is completely neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample My calculation h2s04 = 98g/mol s03 = 80g/mol 98/2 = 49 80/2 = 40 26.7ml x 0.4n = 10.68ml - 0.01067L x/49 + x/40 (0.5-...
*March 31, 2013*

**chemistry (Pls Bob check for me)**

A 0.5g of fuming h2so4 oleum is diluted with water. This solution is completedly neutralized by 26.7ml of 0.4N naoh. Find the percentage of free so3 in the sample The asnwer 20.6% My calculation h2so4= 98/2 = 49 so3= 80/2 = 40 26.7ml x 0.4N = 10.68ml - 0.01068L x/49 + x/40 (0....
*March 24, 2013*

**chemistry (Pls Bob check for me)**

A 150mg/dl glucose solution was diluted 1:2. Then 0.5ml of it was added to 4.5ml saline. 2ml of this solution was diluted to 10ml. What is the final concentration of glucose? 1L = 10dL 1dL = 100mL My own calculation 150mg/dL = 150mg/100ml =1.5 150mg/100ml x 1/2 x 1/10 x 1/5 1....
*March 16, 2013*

**chemistry (Pls Bob check for me)**

A 150 mg/dl glucose solution was diluted 1:2. Then 0.5ml of it was added to 4.5ml saline 2 ml of this solution was diluted to 10ml. What is the final concentration of glucose? 1L= 10dl 1dl =100mL Answer 0.015mg/dl
*March 16, 2013*

**chemistry**

The formula weight of an acid is 82. In a titration, 100cm3 of the solution of this acid containing 39g of this acid per litre were completely neutralized by 95 cm3 of aqueous naoh containing 40g of naoh per litre. What is the basicity of this acid? Bob helps me
*March 10, 2013*

**chemistry**

0.5g of fuming h2s04 oleum is diluted with water. This solution is completely neutralized by 26.7ml of 0.4N naoh. Find the percentage of free s03 in the smaple solution. The answer 20.6% Assume that 50ml of 0.01M edta solution were added to 25ml of solution and the excess edta...
*March 3, 2013*

**chemistry**

Find the concentration of calcium in ppm in a 3.5g pill that contains 40.5mg of ca? Answer 1.16ppm Bob helps me
*March 2, 2013*

**chemistry**

Assume that 50ml of 0.01M edta solution were added to 25ml of co solution and the excess edta required 20ml of 0.005M mg solution. Calculate ppm co+ present. The answer 942ppm Bob helps me
*March 2, 2013*

**chemistry (Pls Bob check for me)**

A 1.2g mixture of na2c03 and k2c03 was dissolved in water to form 100cm3 of s solution, 20cm3 of this solution required 40cm3 of 0.1N hcl for neutralization. Calculate the wieght of na2c03 and k2c03 in the mixture The answer 0.4G My own calculation: na2c03 weight = 106g/2 = ...
*February 24, 2013*

**chemistry**

0.50g of mixture of k2c03 and li2c03 requires 30ml of a 0.25N HCL solution for neutralization. What is the percentage composition of the mixture? The answer k2c03 96% The answer li2c03 4% I want Bob step by step explain. I do not still understand x/69 + y/37 = 0.0075 How to ...
*February 24, 2013*

**chemistry**

0.50g of a mixture of K2c03 and Li2c03 requires 30ml of a 0.25N hcl solution for neutralization. What is the percentage composition of the mixture? The answer k2c03 96% Li2c03 4% K2c03 weight 138g/mol 138/2 = 69 Li2c03 weight 74g/mol 74/2 = 37 30ml x 0.25N = 7.5 / 1000 x= 0....
*February 22, 2013*

**chenistry**

25ml of a solution of na2c03 having a sepcific gravity of 1.25g solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N h2s04 that will be compltetly neutralized by 125g of na2c03 solution. In first part I understand ...
*February 17, 2013*

**chemistry**

25ml of a solution of na2c03 having a specific gravity of 1.25g ml required 32.9ml of a solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N heso4 that will be compltetly neutralized by 125g of na2c03 solution The ...
*February 17, 2013*

**chemistry**

25ml of a solution of na2c03 having a specific gravity of 1.25g ml required 32.9ml of a solution of hcl containing 109.5g of the acid per litre for complete neutralization. Calculate the volume of 0.84N Hes04 that will be completely neutralized by 125g of nac03 solution The ...
*February 16, 2013*

** chemistry**

A definite amount of NH4cl was boiled with 100ml of 0.8N naoh for complete reaction . After the reaction mixture containing excess of naoh was neutralized with 12.5ml of 0.75N NH2so4. Calculate the amount of NH4cl taken. I want you to explain about 100ml x 0.8N = 80 12.5ml x 0...
*February 10, 2013*

**chemistry**

A definite amount of nh4cl was boiled with 100ml of 0.8N naoh for complete reaction. After the reaction, the reactant mixture containing excess of naoh was neutralized with 12.5ml of 0.75N nh2s04. Calculate the amount of nh4cl takedn The answer 3.78g who helps
*February 10, 2013*

**chemistry**

In the previous example, how many ml of 0.1M kmn04 are required to react with 30ml of 0.05M h202? the answer 3ml who helps
*February 10, 2013*

**chemistry**

A definite amount of nh4cl was boiled with 100ml of 0.8N naoh for complete reaction. After the reaction, the reaction mixture containing excess of naoh was neutralized with 12.5ml of 0.75N nh2s04. Calculate the amount of nh4cl taken. The answer 3.78g who helps me
*February 10, 2013*

**chemistry**

A definite amount of NH4cl was boiled with 100ml of 0.8N naoh for complete reaction. After the reaction, the reactant mixture containing excess of naoh was neutralized with 12.5ml of 0.75N nh2s04. Calculate the amount of nh4cl taken. The answer 3.78g who helps
*February 10, 2013*

**chemistry**

How many grams of kmn04 are contained in 35ml of 0.05N kmn04 used in the following reaction in basic solution. the answer 0.0922g who helps
*February 9, 2013*

**chemistry**

calculate the number of g fes04 that will be oxidized by 24ml of a 250N Kmn04 in a solution acidified with sulfuric acid. The unbalanced equation for the reaction. The answer 0.912g Who helps
*February 9, 2013*

**chemistry**

80ml of hcl is added to 2.5gm of pure cac03 when the reaction is over, then 0.5gm cac03 is left. Find the normality of the acid The answer 0.5N Who helps step by step
*February 3, 2013*

**chemistry**

How many ml of 0.05M kmn04 acidic are diluted to oxidize 2gm of fes04 is diluted solution The answer 52.63 ml I want you step by step clearly explain Thank you for Bob
*January 27, 2013*

**chemistry**

What weight of na2co3 of 95% purity would be required to neutralize 45.6ml of 0.235n acid? I want you step by step clearly explain answer 0.5978gm
*January 27, 2013*

**chemistry**

5ml of 8N Hn03, 4.8ml of 5N hcl and a certain volume of 17M h2s04 are mixed together and made up to 2 liter. 30ml of this acid mixture exactly neutralizes 42.9ml of na2c03, 10 h20 in 100ml of water .Calculate the amount of sulphate ion in gm present in solution Who help step ...
*January 27, 2013*

**chemistry**

5ml of 8N HN03, 4.8ml of 5N HCL and a certain volume of 17M H2s04 are mixed together and made up to 2 liter 30ml of this acid mixture exactly neutralize 42.9ml of na2c03, 10 H2O in 100ml of water. Calculate the amount of sulphate 10m in gm present in solution The 6.5gm who help
*January 27, 2013*

**chemistry**

what weight of na2c03 of 95% purity would be required to neutralize 45.6ml of 0.235N acid? Who Helps
*January 27, 2013*

**chemistry**

In the previous example, how many ml of 0.1M KMN04 are required to react with 30ml of 0.05M H202? The answer 3ml Who helps
*January 13, 2013*

**chemistry**

In the previous example, how many ml of 0.1M KMN04 are required to react with 30ml of 0.05M H2o2? The answer 3ml who helps
*January 13, 2013*

**chemistry**

calculate the molarity of H2O2 if 25ml of H2O2 required 12 ml of 0.1M KMN04? The right answer 0.24M who helps
*January 13, 2013*

**difficult chemistry**

502g sample of dry cac03 and cacl2 mixture was dissolved in 25ml of 0.925 M HCL solution. What was cacl2 percentage in original sample if 22ml of 0.09312 m NAOH was used to titrated excess HCL? The answer 29.89% Who helps me
*January 12, 2013*

**difficult chemistry**

Calculate the molarity of thiosulfate solution if 0.400mg of Kl03 required 100ml of the thiosulfate solution provided that excess kl and hcl were added. Who helps me
*January 12, 2013*

**difficult chemistry**

100 Liters of air at STP is slowly bubbled through 200ml of 0.03N Ba(OH)2 solution. The Baco3 formed due to reaction is filtered and few drops of phenolphthalein is added to the solution rendering it pink. The solution required 25ml of 0.2N HCL solution when indicator turned ...
*January 12, 2013*

**chemistry**

8.058g x 10-2 kg of washing soda na2co3 10h20 is dissolved in water to obtain 1L of a solution of density 1077.2kg/m3. Calculate the molarity, molality and mole fraction of na2c03 in the solution? who helps
*January 1, 2013*

**chemistry**

10g of nacl were added to 10ml of water. the resulting solution was diluted to be a final volume of 250ml. Given the density of the solution is 0.9864g/ml. Calculate the molality of the solution. Who helps
*December 25, 2012*

**chemistr**

10g of nacl were added to 10ml of water. he resulting solution was diluted to be a final volume of 250ml. Calculate the molarity of the solution. My calculation 10g/58g/mol = 0.173M 0.173M/025L =0.689M Given the density of the solution is 0.9864g/ml. Calculate the molality of ...
*December 25, 2012*

**chemistry**

6.85ml of 0.015M SN2+
*December 16, 2012*

**chemistry**

A 1.252g sample containing iron is dissolved and converted to FE 2+ followed by addition of 25ml of 0.01M k2 cr2 07. The excess K2cr207 required 6.85 ml of 0.015M. Find the percentage FE304 FW 231g/mol in the sample. The right 7.96% Who helps me
*December 16, 2012*

**chemistry**

An aqueous solution is 0.273m kcl. What is the molar concentration of potassium chloride, kcl? The density of the solution is 1.011 x 1 1000 g/L Who helps me
*December 16, 2012*

**chemistry**

A 1.252g sample containing iron dissolved and converted to Fe followed by ADDITION OF 25ML OF 0..01m K2CR2 07. THE EXCESS K2CR207 REQUIRED 6.85ML OF 0.015m SN2+. Find the percentage fe3 04 fw 213g/mol in the sample. Who helps me
*December 11, 2012*

**chemistry**

A 0.200g sample containing mn02 was dissolved and analyzed by addiction of 50ml of 0.100M fe to drive the reation. The excess fe required 15ml of 0.0200 m kmn04. Find % mn304 fw 228.8mg/mmol in the sample? who helps me
*December 9, 2012*

**chemistry**

A 0.4671g sample containing sodium bicarbonate was titirated with hcl requiring 40.72ml. The acid was standardized by titrating 0.1876g of sodium carbonate FW 106mg/mmol requiring 37.86ml of the acid. Find the percentage of nahc03 fw 84mg/mmol in the sample. My own calculation...
*November 25, 2012*

**chemistry**

A 0.682g sample of impure Na2hco3 yield a solid residue consisting of nahco3 and other solid a mass of 0.467g. What was the mass percent of nac03in sample. I do not understand I want you step by step. Clear explain for me. nahco3 = 84 n22hco3 = 106
*November 23, 2012*

**chemistry**

A 0.682g sample of impure NAHco3 yielded a solid residue consisting of NAc03 and other solid with a mass of 0.467g. What was the mass percent of NAc03 in the sample. The right 85.4% Who help
*November 23, 2012*

**chemistry**

A 0.46771g sample containing sodium bicarbonate was titrated with hcl requiring 40.72ml. The acid was standardized by titrating 0.1876g requiring 37.86ml of the acid.Find the percentage of NAHCo3 FW 84mg/mmoml in the sample? Who helps me to solve it
*November 22, 2012*

**Chemistry Math**

A 0.4671g sample containing sodium bicarbonate was titrated with HCL requiring 40.72ml. The acid was standardized by titrating 0.1876g of sodium carbonate FW 106mg/mmol requiring 37.86ml of the acid. Find the percentage of NAHO3 FW 84mg/mmol in the sample Who helps me to solve...
*November 18, 2012*

**Chemistry**

A 5g mn sample was dissolved in 100ml water. If the percentage of M (At wt =55g/mol) in the sample is about 5%. What volume is needed to prepare 100ml of approximately 3.0 x 10-3 M solution Who helps for me. The right answer 6.7ml
*November 4, 2012*

**Chemistry**

Find the number of mg Na2co3 FW = 106g/mol required to prepare 500ml of 9.2ppm NA solution. ppm = mg/L 9.2mg/1000ml x= 0.0092 x 500ml = 4.6mg x 23 = 106mg The right answer 10.6mg NA weight = 23 Who helps me to solve it for me.
*November 4, 2012*

**chemistry**

A solution of ethanol (C2H50H) in water is prepared by dissolving 76.9ml of ethanol (density0.79g/cm3) in enough water to make 0.25L of solution. What is the molarity of ethanol in this solution Who helps me
*September 29, 2012*

**chemistry**

Concentrated hydrochloric acid is a 38% solution of hcl in water and has a density of 1.18g/ml. How many milliliters of concentrated hydrochloric acid are needed to prepare 500ml of a 1:200 hcl solution. My own three calculation 1.18g/ml x 1000ml x0.38 / 36.46 = 12.3M 500ml x ...
*September 20, 2012*

**chemistry**

Assuming the density of a 5% acetic acid solution is 1g/ml, determine the volume of the acetic acid solution necessary to neutralize 25ml of 0.1m naoh also record this calculation on your report My calculation 1g/ml x 1000ml x 0.05/61 = 0.81M 0.81m = 25ml x 0.1 0.81m = 2.5 x= ...
*September 9, 2012*

**chemistry**

A flask containing 5ml of 3M hcl solution required 14.45ml of 1M NAOH for titration. How many moles of hcl are present in the solution. The density of 3M hcl is 1.05g/ml My calculation 0.005 x 3 = 0.015m 0.01445 x 1 = 0.01445m Who helps me to solve it
*September 2, 2012*

**chemistry**

A flask containing 5.00ml of 3M hcl solution required 14.45 of 1.00M naoh for titration. How many moles of hcl are present in the solution? The density of 3M hcl is 1.05g/ml. My calculation 0.005L x 3M = 0.015 moles 0.01445 x 1m = 0.01445 moles Who helps me to solve it
*September 2, 2012*

**chemistry**

What is the molarity of 100ml of ethanol dissolved in 1.5L of water? Ethanol has a specific of 0.789, and its molecular weight is 46g/mol. My calculation 0.789 x 1.5L = 1.1184 1.184/46g = 0.025 x= 26M I am not sure answer. Pls help
*August 19, 2012*

**Chemistry**

Concentrated hydrochloric acid is a 38% solution of hcl in water and has a density of 1.18g/ml. How many milliliters of concentrated hydrochloric acid are needed to prepare 500ml of 1:200 hcl solution. My calculation 1.18g/ml x 1000ml x 0.38/36.46 = 12.3M 12.3M = 500ml x 1:200...
*August 19, 2012*

**chemistry**

A 0.446g sample of an unknown monoproctic acid was titrated with 0.105 M KOH The molar mass is 120g/mol. What is the PKA value? Pls helps me again
*August 19, 2012*

**Chemistry**

What is the mass/ vol% ethanol in an ethanol water solution with density of 0.875g/ml and containing 65% ethanol by volume? The density of pure ethanol is 0.789g/ml? My calculation 0.875g/ml x 1000ml x 0.65/ 46.06 = 12.34m 12.34/0.789g = 15.6g I am not sure answer. Pls help
*August 18, 2012*

**chemistry**

A 0.446g sample of an unknown monoprotic acid was titrated with 0.105M KOH the molar mass is 120g/mol. What is the pka value? My calculation 0.105M x 0.446g = 0.04683 x 1000 =46.83 46.83 x 120 = 56196 changes 5.6gm This answer 5.6gm I am not sure answer. Pls help
*August 12, 2012*

**Chemistry**

Suppose a student started with 129mg transcinnamic acid and 0.50ml of 10% bromine solution and after the reaction and workup ended up with 0.216g of brominated product. Calculate the student's theoretical and percent yield. My calculation 0.50ml x 10% = 5ml And then 0.126g...
*August 12, 2012*

**chemistry**

A solution of sodium cyanide nacn has a ph of 12.10. How many grams of nacn are in 425ml of solution with the same ph? I do not understand how to calculate this question. Who helps me to solve it for me. Thank you. The right answer 209gm. A 0.446g sample of an unknown ...
*August 11, 2012*

**chemistry**

How much ethanol C2H50H in liters d = 0.789g/ml must be dissolved in water to produce 190.5 of 1.65M C2H50H? Molarity problem. My calculation 0.789g/ml x 1000ml/46.06 = 17.12M 17.12M = 190.05 x 1.66 17.12m = 314.325 x= 314.325/17.12 x=18.36 I am not sure answer. Please help me...
*August 11, 2012*

**chemistry**

Concentrated hydrochloric acid has specific gravity 1.18 and 36% hcl weight. What volume of concentrated hcl must be diluted to give 50gm of 10 percent My calculation 50gm x 10% = 5gm 5gm x 36.46 =182.3/ 0.36 = 506 / 1.18 = 429 ml I am not sure answer. Who helps me to solve it...
*August 11, 2012*

**Chemistry**

A solution of is prepared by adding 50.3ml of concentrated hydrochloric acid and 16.6ml of concentrated nitric acid to 300ml of water. More water is added until the final volume is 1.00L. Calculate (H+) (OH) and the PH for this solution. Hint concentrated HCL is 38% hcl by ...
*August 11, 2012*

**chemistry**

How to calculate PH 2.13 = 0.007M Why? Please show for me. I want you step by step for me.
*August 11, 2012*

**chemistry**

How to calculate PH 1.9 =0.0126M I do not still understand. I want you step by step to calculate for me. Thank you.
*August 10, 2012*

**chemistry**

Concentrated hydrochloric acid is a 38% solution of hcl in water and has a density of 1.18g/ml. How many milliliters of concentrated hydrochloric acid are needed to prepare 500ml of a 1:200 hcl solution? My calculation 500ml x 1:200 =2.5 1:200 means 1gm/200ml 2.5 x 36.5 = 91....
*August 9, 2012*

**chemistry**

Concentrated hydrochloric acid has specific gravity of 1.18 and 36% hcl weight. What volume of concentrated hcl must be diluted to give 50 gm of 10 percent. My calculation 50gm x 0.1 = 5gm 5gm x 36.5 = 182.5 182.5 / 0.36 = 506.9 506.9 / 1.18 = 430 ml This is answer. I am not ...
*August 9, 2012*