• HARD CHEMISTRY HELP!: IN which direction (left to right or right to left) will the following reactions occur? Sn + Ag^+ ---- Sn^2+ + Ag?

    Look at the reactivity series. Sn is above Ag in the series. Therefore Sn will displace Ag+ . The equation is Sn(s) + 2Ag^+ (aq) → Sn^2+ (aq) + 2Ag(s)
    Look at the reactivity series. Sn is above Ag in the series. Therefore Sn will displace Ag+ . The equation is Sn(s) + 2Ag^+ (aq) → Sn^2+ (aq) + 2Ag(s)
    1 answer · Chemistry · 9 hours ago
  • Ammonium carbonate and calcium carbonate how to tell these two substances apart?

    Ammonium carbonate is soluble in water Calcium carbonate is insoluble in water.
    Ammonium carbonate is soluble in water Calcium carbonate is insoluble in water.
    1 answer · Chemistry · 11 hours ago
  • Wasting food?

    If the food has been stored in a frozen state inside a freezer , then there is no reason for the food to be unsafe for consumption.
    If the food has been stored in a frozen state inside a freezer , then there is no reason for the food to be unsafe for consumption.
    8 answers · Diet & Fitness · 24 hours ago
  • Chemistry:In this pH question,I am confused on a specific part given in the mark scheme so can anyone clarify how this is worked out,better?

    Best answer: You can calculate the pH of the buffer solution using the Henderson - Hasselbalch equation: pH = pKa + log ( [NaY] / [HY] ) You calculate pKa of the acid pKa = -log Ka = -log (1.35 × 10–5) pKa = 4.87 Now to answer you query: When you write [NaY] you are saying : molar concentration of NaY in mol per dm³ solution. You know that you... show more
    Best answer: You can calculate the pH of the buffer solution using the Henderson - Hasselbalch equation: pH = pKa + log ( [NaY] / [HY] ) You calculate pKa of the acid pKa = -log Ka = -log (1.35 × 10–5) pKa = 4.87 Now to answer you query: When you write [NaY] you are saying : molar concentration of NaY in mol per dm³ solution. You know that you have dissolved 0.0236 mol of the salt NaY in 50.0 cm3 solution How many mol have you dissolved in 1dm³ of solution You know that there are 1000cm³ in 1dm³ . Therefore mol in 1dm³ solution = 0.0236*1000/50 = 0.472 mol /dm³ You already know that [HY] = 0.428 mol / dm³ Now substitute into the equation pH = 4.87 + log ( 0.472/ 0.428 ) pH = 4.87 + log 1.102 pH = 4.87 + 0.04 pH = 4.91
    2 answers · Chemistry · 2 days ago
  • What i the difference between Ferrous Fumarate and Ferrous Sulphate (UK)?

    Iron(II) fumarate, also known as ferrous fumarate, is the iron(II) salt of fumaric acid, Ferrous fumarate has the chemical formula C4H2FeO4. This compound has nothing to do with ferrous sulphate , which is FeSO4. You should see that Ferrous fumarate cannot be confused with ferrous sulphate – it does not contain any sulphur.
    Iron(II) fumarate, also known as ferrous fumarate, is the iron(II) salt of fumaric acid, Ferrous fumarate has the chemical formula C4H2FeO4. This compound has nothing to do with ferrous sulphate , which is FeSO4. You should see that Ferrous fumarate cannot be confused with ferrous sulphate – it does not contain any sulphur.
    3 answers · Other - General Health Care · 2 days ago
  • What would be the partial pressure of N2 in a 50 degrees Celsius which there is 0.20 moles N2 and 0.10 moles CO2 at apressure of 101.3 kPa?

    Temperature is not important Total moles of gas = 0.2 + 0.1 = 0.3 mol Partial pressure of N2 = moles N2 / total moles * pressure Partial pressure N2 = 0.2/0.3 * 101.3kPa Partial pressure N2 = 67.5kPa To correct significant digits: Partial pressure N2 = 68kPa
    Temperature is not important Total moles of gas = 0.2 + 0.1 = 0.3 mol Partial pressure of N2 = moles N2 / total moles * pressure Partial pressure N2 = 0.2/0.3 * 101.3kPa Partial pressure N2 = 67.5kPa To correct significant digits: Partial pressure N2 = 68kPa
    3 answers · Chemistry · 2 days ago
  • Chemistry question wants to determine density for following question.?

    What units do you want to use for the density . It is usual in science to quote density as g/mL or kg/m³ Density is defined as : mass per unit volume . In practice you can use any units you wish - but this may not be of value to others. You do not specify what you want as units , but because you have mass in pounds and volume in mL, I will use... show more
    What units do you want to use for the density . It is usual in science to quote density as g/mL or kg/m³ Density is defined as : mass per unit volume . In practice you can use any units you wish - but this may not be of value to others. You do not specify what you want as units , but because you have mass in pounds and volume in mL, I will use these units Density = 0.255lb / 132.8mL Density = 0.0019lb/mL
    3 answers · Chemistry · 2 days ago
  • Need help with this significant figure problem for chemistry?

    I agree with your answer.
    I agree with your answer.
    3 answers · Chemistry · 2 days ago
  • If you pee in a container of bleach, would it make mustard gas?

    No. My answer would be similar to that submitted by Pisgachemist. I do not see any reason to repeat this totally correct information.
    No. My answer would be similar to that submitted by Pisgachemist. I do not see any reason to repeat this totally correct information.
    6 answers · Chemistry · 3 days ago
  • 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 · 4 days 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 · 5 days 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 · 5 days 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 · 6 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 · 6 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 · 6 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 · 1 week 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 · 1 week 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 · 1 week 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 · 1 week 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 · 1 week ago