• Chem solution question, pls help?

    I continue to be puzzled by the strange density values given frequently in Y!A. I do not know where these density values come from . But they make no logic: Read the data given and some of the solutions calculated 35g NaOH dissolved in 1.5x10^2g water, That is 35g NaOH dissolved in 150g or 150mL water It is logical to expect the volume of the... show more
    I continue to be puzzled by the strange density values given frequently in Y!A. I do not know where these density values come from . But they make no logic: Read the data given and some of the solutions calculated 35g NaOH dissolved in 1.5x10^2g water, That is 35g NaOH dissolved in 150g or 150mL water It is logical to expect the volume of the final solution to be in excess if 150mL But we are told: Density = 1.57g/mL By calculation , the volume is 118 mL How is it possible that you take 150g or 150mL of water , add 35g of NaOH to this water and then the total volume is 118mL ? There is no logic in this question - or unhappily the answers submitted. I have checked my density data for NaOH solutions. We have 35g NaOH in a total of 185g solution = 18.9%m/m Density of NaOH solutions 19.16%m/m: density = 1.210g/mL 18.26%m/m: density = 1.200g/mL We can approximate the density of a 18.9%m/m NaOH solution to be 1.208g/mL Now do the calculation 185 g / 1.208 g/mL = 153.1mL ( which is what we can expect) 35 g of NaOH = 0.875 mole molarity =moles / volume in L Molarity = 0.875 mol / 0.1531L Molarity = 5.7mol /L or 5.7M Do some “ crude “ chemistry: You have 35g NaOH in 150mL water This is APPROXIMATELY 35*1000/150 = 233g NaOH /L Mol NaOH / L = 233/40 = 5.8mol /L Which allowing for the “crude” calculation: Checks my above reasoning. Incidentally , an NaOH solution with density = 1.530 g/mL has a concentration of 50.50%m/m This density is less than the submitted 1.57g/mL
    3 answers · Chemistry · 15 hours ago
  • P4O6 + Water....Why does the reaction end up with H3PO3 and not H3PO4?

    P4O6 is the anhydride of phosporous acid H3PO3 You can write a balanced equation for this reaction : P4O6 + 6H2O → 4H3PO3 But P4O6 does not react with water to produce H3PO4 You cannot balance this equation: P4O6 + H2O → H3PO4 This is similar to the oxide of sulphur: SO2 It reacts with water to produce H2SO3 - Not H2SO4 show more
    P4O6 is the anhydride of phosporous acid H3PO3 You can write a balanced equation for this reaction : P4O6 + 6H2O → 4H3PO3 But P4O6 does not react with water to produce H3PO4 You cannot balance this equation: P4O6 + H2O → H3PO4 This is similar to the oxide of sulphur: SO2 It reacts with water to produce H2SO3 - Not H2SO4
    2 answers · Chemistry · 1 day ago
  • Which of the following do not conduct electricity - 1. Carbon Rod 2. Zinc Rod 3. Brass Rod 4. Iron Rod Please help.?

    All your choices conduct electricity.
    All your choices conduct electricity.
    6 answers · Chemistry · 1 day ago
  • Astudent diss 28.3g of sodium phosphate to prepare a 4.29L sol 2nd student dilutes this sol to 5.54L If completely dissociates find #of mols?

    It is a courtesy to other contributors who may wish to assist you to submit your questions in a readable way . Do not use abbreviations such as diss , sol and # - these can mean different things to different people. Use correct punctuation and complete sentences. Despite what I have said above I will try and help you. I have to guess that you... show more
    It is a courtesy to other contributors who may wish to assist you to submit your questions in a readable way . Do not use abbreviations such as diss , sol and # - these can mean different things to different people. Use correct punctuation and complete sentences. Despite what I have said above I will try and help you. I have to guess that you want to know the number of mol of Na3PO4 in 28.3g of compound. Dissolving and diluting the solution does not change the number of mols of solute Molar mass of Na3PO4 = 163.94 g/mol Mol Na3PO4 in 28.3g = 28.3g/ 163.94g/mol = 0.172624 mol Answer to correct significant figures = 0.173 mol Na3PO4 I have read your comment: To answer what you ask: Na3PO4 dissociates Na3PO4(aq) → 3Na+ (aq) + PO4 3-(aq) 1mol Na3PO4 produces 3 mol Na+ ions and 1 mpol PO4 3- ion You have 0.173 mol Na3PO4 in solution Therefore you have 3 * 0.173 = 0.519 mol Na+ ions 0.173 mol PO4 3- ions But I wonder if this is what the question asks for? You will notice that I have not used any of the dilution data given . Why was this data included if it is not used . Have you submitted the question EXACTLY AND COMPLETELY as it was given to you? Based on what I can read and how questions are normally set , I wonder if the complete question is not similar to: A student dissolves 28.3 g of sodium phosphate to prepare a 4.29 L solution. Another student then dilutes ( X mL -you fill in the volume) of this solution to 5.54 L. If sodium phosphate completely dissociates, determine the molar concentration of each ion in the final solution. Check the original question and advise
    3 answers · Chemistry · 2 days ago
  • CHEM HELP PLS?

    Molar mass of KOH is 56.10 g/mol Mol KOH in 4.25g = 4.25/56.10 = 0.0758 mol From the balanced equation 1 mol Cl2 will react with 2 mol KOH 0.0758/2 = 0.0379 mol Cl2 will react with 0.0758 mol KOH At the STP conditions specified ( 0°C and 101.325 kPa pressure) 1 mol Cl2 has volume = 22.4L 0.0379 mol Cl2 has volume = 0.0379 *22.4 = 0.849L Cl2... show more
    Molar mass of KOH is 56.10 g/mol Mol KOH in 4.25g = 4.25/56.10 = 0.0758 mol From the balanced equation 1 mol Cl2 will react with 2 mol KOH 0.0758/2 = 0.0379 mol Cl2 will react with 0.0758 mol KOH At the STP conditions specified ( 0°C and 101.325 kPa pressure) 1 mol Cl2 has volume = 22.4L 0.0379 mol Cl2 has volume = 0.0379 *22.4 = 0.849L Cl2 gas If you want a more formal solution: Use equation PV = nRT 1*V = 0.0379 * 0.082057*273.15 V = 0.849L Cl2 gas
    3 answers · Chemistry · 2 days ago
  • Sig fig substraction help!? chem?

    WARNING: the rule for add/subtract are different from multiply/divide. A very common student error is to swap the two sets of rules. Another common error is to use just one rule for both types of operations For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the numbers ONLY. Here is what to... show more
    WARNING: the rule for add/subtract are different from multiply/divide. A very common student error is to swap the two sets of rules. Another common error is to use just one rule for both types of operations For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the numbers ONLY. Here is what to do: 1) Count the number of significant figures in the DECIMAL portion of each number in the problem. (The digits to the left of the decimal place are not used to determine the number of decimal places in the final answer.) 2) Add or subtract in the normal fashion. 3) Round the answer to the LEAST number of places in the decimal portion of any number in the problem. In you problem: 80 has no figures in the decimal portion of the number 2.99 has 2 figures in the decimal portion of the number Therefore the answer must have no figures in the decimal portion of the number Conventionally: 80 - 2.99 = 77.01 To correct number of significant figures: Answer = 77
    3 answers · Chemistry · 2 days ago
  • Can someone please help me with steps?

    heat energy = mass X specific heat X change in temp 2105 = 42.02 X SH X ( 122.1 - 24.9) SH = 2105 / ( 42.02 X 97.2 ) SH = 0.515J/g°C
    heat energy = mass X specific heat X change in temp 2105 = 42.02 X SH X ( 122.1 - 24.9) SH = 2105 / ( 42.02 X 97.2 ) SH = 0.515J/g°C
    2 answers · Chemistry · 3 days ago
  • I'm NO nearer the answer to mine:- Can a Public group be, later, restricted?

    Why do you say that you did not have enough room in your original question? Yahoo gives you adequate room if you use the system correctly. Click on the details tab Click on the " expand" option. You have a box marked : Question. Here you enter a short question title . In your case : Can a public group be restricted Then immediately... show more
    Why do you say that you did not have enough room in your original question? Yahoo gives you adequate room if you use the system correctly. Click on the details tab Click on the " expand" option. You have a box marked : Question. Here you enter a short question title . In your case : Can a public group be restricted Then immediately under this you will find a large box marked: Details. Here you enter all the details about what you want to ask . In you case : "They" said that restricted, private, etc. groups could not be made "Public" later. ("Restricted and Private groups cannot be made Public later") SO:- I want to know if the opposite is true. Can a "PUBLIC" group, later, not be made into a "RESTRICTED" group. You will have plenty of space to add far more details if necessary . You do this whenever you have a question to submit to Yahoo. There is no restriction from space available.
    4 answers · Yahoo Groups · 3 days ago
  • Break even quantity exam question?

    Let the break even quantity be X 60,000 + 20X = 69000 + 15X 5X = 9000 X = 1800 units
    Let the break even quantity be X 60,000 + 20X = 69000 + 15X 5X = 9000 X = 1800 units
    3 answers · Mathematics · 3 days ago
  • What is the 2 refering to in trans 2 butene?

    Butene is a 4 carbon length molecule that contains a double bond between two of the carbons You can have: C=C-C-C You see the double bond is between C1 and C2 . This is named 1 butene Or you can have : C-C=C-C . Here the double bond is between C2 and C3 . This is 2 butene. I assume you know what trans means because you do not ask for this.... show more
    Butene is a 4 carbon length molecule that contains a double bond between two of the carbons You can have: C=C-C-C You see the double bond is between C1 and C2 . This is named 1 butene Or you can have : C-C=C-C . Here the double bond is between C2 and C3 . This is 2 butene. I assume you know what trans means because you do not ask for this. Incidentally - I hope that you have sufficient knowledge to know that you cannot have 3 butene .
    3 answers · Chemistry · 4 days ago
  • Chem help?

    The molar mass of water is 18g/mol You have 180g water - which is 10 mol water You know that 1 mol water conains exactly the same number of molecules as 1 mol glucose In order to have 1 molecule of glucose for every 10 molecules of water , you must therefore use 1 mol glucose added to the 10 mol water. Mass of 1 mol glucose = 6*12 + 12*1 + 6*16... show more
    The molar mass of water is 18g/mol You have 180g water - which is 10 mol water You know that 1 mol water conains exactly the same number of molecules as 1 mol glucose In order to have 1 molecule of glucose for every 10 molecules of water , you must therefore use 1 mol glucose added to the 10 mol water. Mass of 1 mol glucose = 6*12 + 12*1 + 6*16 = 180g/mol Therefore you must add 180g glucose to the 180g water.
    1 answer · Chemistry · 4 days ago
  • Challenge: gas law question?

    There is no reaction between the SO2 and Ar. When the valve is opened, you have a single container of mixed gases at STP. If we accept that the volume remains constant after the change in temperature , we can calculate the final pressure using P1/T1 = P2/T2 P1 = 100kPa T1 = 25°C = 298K P2 = ?? T2 = 70°C = 343K 100/298 = P2/343 P2 =... show more
    There is no reaction between the SO2 and Ar. When the valve is opened, you have a single container of mixed gases at STP. If we accept that the volume remains constant after the change in temperature , we can calculate the final pressure using P1/T1 = P2/T2 P1 = 100kPa T1 = 25°C = 298K P2 = ?? T2 = 70°C = 343K 100/298 = P2/343 P2 = 343*100*298 P2 = 115 kPa.
    2 answers · Chemistry · 4 days ago
  • Can someone please help me with this challenging gas law question?

    Best answer: You have two containers with the same pressure ( SATP) You connect these containers by opening the valve. The pressure will not change . Final pressure is 100kPa - which is the standard SATP pressure. The differences in volume ( or contents) of the two containers is not of any consequence.
    Best answer: You have two containers with the same pressure ( SATP) You connect these containers by opening the valve. The pressure will not change . Final pressure is 100kPa - which is the standard SATP pressure. The differences in volume ( or contents) of the two containers is not of any consequence.
    1 answer · Chemistry · 4 days ago
  • What is the boiling point in celsius of a 1.56m aqueous solution of K3PO4?

    K3PO4 dissociates: K3PO4(aq) → 3K+(aq) + PO4 3-(aq) i = 3+1=4 Boiling point increase = Kb * i * m Boiling point increase = 0.512 * 4 * 1.56 Boiling point increase = 3.195°C Boiling point of solution = 103.195°C
    K3PO4 dissociates: K3PO4(aq) → 3K+(aq) + PO4 3-(aq) i = 3+1=4 Boiling point increase = Kb * i * m Boiling point increase = 0.512 * 4 * 1.56 Boiling point increase = 3.195°C Boiling point of solution = 103.195°C
    3 answers · Chemistry · 4 days ago
  • How to write equations for electrolysis?

    Best answer: The reactions at each electrode are called half equations. The half equations are written so that the same number of electrons occur in each equation. 2Na+ + 2e- → 2Na (sodium metal at the (-)cathode). 2Cl- - 2e- → Cl2 (chlorine gas at the (+)anode). Sodium ions gain electrons (reduction) to form sodium atoms. Chloride... show more
    Best answer: The reactions at each electrode are called half equations. The half equations are written so that the same number of electrons occur in each equation. 2Na+ + 2e- → 2Na (sodium metal at the (-)cathode). 2Cl- - 2e- → Cl2 (chlorine gas at the (+)anode). Sodium ions gain electrons (reduction) to form sodium atoms. Chloride ions lose electrons (oxidation) to form chlorine atoms. The chlorine atoms combine to form molecules of chlorine gas. The overall reaction is 2Na+Cl-(l) → 2Na(s) + Cl2(g)
    2 answers · Chemistry · 4 days ago
  • A chemist burns 750.0 g of C4H8S2 in excess oxygen, according to the following reaction C4H8S2 + 9O2 → 4CO2 + 4H2O + 2SO3?

    Molar mass of C4H8S2 is 120.2363 g/mol Mol in 750.0g = 750.0/120.2363 = 6.238 mol 1mol C4H8S2 produces 2 mol SO3 6.238 mol will produce 2*6.238 = 12.476 mol SO3 Molar mass SO3 = 32+3*16 = 80.0g/mol Theoretical yield = 12.476*80.0 = 998.08g Actual yield is 95.4% Actual mass produced = 95.4/100*998.08 = 952.2g SO3 produced. show more
    Molar mass of C4H8S2 is 120.2363 g/mol Mol in 750.0g = 750.0/120.2363 = 6.238 mol 1mol C4H8S2 produces 2 mol SO3 6.238 mol will produce 2*6.238 = 12.476 mol SO3 Molar mass SO3 = 32+3*16 = 80.0g/mol Theoretical yield = 12.476*80.0 = 998.08g Actual yield is 95.4% Actual mass produced = 95.4/100*998.08 = 952.2g SO3 produced.
    1 answer · Chemistry · 4 days ago
  • A compound has an empirical formula of C2H3O3. If its molar mass is 375.25 g, what is its molecular formula?

    formula mass C2H3O3 = 2*12 + 3*1 + 3*16 = 75g/formula unit How many formula units in molar mass: 375/75 = 5 Therefore molecular formula = ( C2H3O3)5 = C10H15O15.
    formula mass C2H3O3 = 2*12 + 3*1 + 3*16 = 75g/formula unit How many formula units in molar mass: 375/75 = 5 Therefore molecular formula = ( C2H3O3)5 = C10H15O15.
    2 answers · Chemistry · 4 days ago
  • Chemist burns 100g of chromium in excess oxygen.Only one product and it has a mass of 192.31g.What is the empirical formula of this product?

    Best answer: Mass of Cr = 100g molar mass Cr = 52g/mol Mol Cr in 100g = 100/52 = 1.923 mol Mass of O = 192.31g - 100g = 92.31g Molar mass O = 16g/mol mol O i9n 92.31g = 92.31/16 = 5.769 mol Divide mol amounts by lowest value: Cr = 1.923/1.923 = 1 O = 5.769/1.923 = 3. Empirical formula = CrO3
    Best answer: Mass of Cr = 100g molar mass Cr = 52g/mol Mol Cr in 100g = 100/52 = 1.923 mol Mass of O = 192.31g - 100g = 92.31g Molar mass O = 16g/mol mol O i9n 92.31g = 92.31/16 = 5.769 mol Divide mol amounts by lowest value: Cr = 1.923/1.923 = 1 O = 5.769/1.923 = 3. Empirical formula = CrO3
    3 answers · Chemistry · 4 days ago
  • Combustion analysis is done on 522.4g of a substance.Produces 499.4g carbon dioxide and 204.5g water.What is empirical formula of substance?

    Mass C in 499.4g CO2 = 12/44*499.4 = 136.2 g Mol C = 136.2/12 = 11.35 mol C Mass H in 204.5g H2O = 2/18*204.5 = 22.72g Mol H i = 22.72/1 = 22.72 Divide by smaller value C = 11.35/11.35 = 1 H = 22.72/11.35 = 2 Empirical formula = CH2
    Mass C in 499.4g CO2 = 12/44*499.4 = 136.2 g Mol C = 136.2/12 = 11.35 mol C Mass H in 204.5g H2O = 2/18*204.5 = 22.72g Mol H i = 22.72/1 = 22.72 Divide by smaller value C = 11.35/11.35 = 1 H = 22.72/11.35 = 2 Empirical formula = CH2
    1 answer · Chemistry · 4 days ago