• How many molecules are in 97.8g of FeCl3?

    FeCl3 is not a molecule, it is an ionic compound. The proper question should be, "How many formula units are there in 97.8 g of FeCl3?". Molar mass FeCl3 = Fe (55.5) + 3Cl (3 x 35.5) = 162 g/mole 97.8 g FeCl3 x (1 formula unit FeCl3 / 162 g FeCl3) = 0.604 formula units FeCl3
    FeCl3 is not a molecule, it is an ionic compound. The proper question should be, "How many formula units are there in 97.8 g of FeCl3?". Molar mass FeCl3 = Fe (55.5) + 3Cl (3 x 35.5) = 162 g/mole 97.8 g FeCl3 x (1 formula unit FeCl3 / 162 g FeCl3) = 0.604 formula units FeCl3
    6 answers · Chemistry · 6 months ago
  • Chemistry help?

    N2 has an effusion rate of 79 mL/s. According to Graham's Law, Rate gas 1 / Rate gas 2 = (MM gas 2 / MM gas 1)^1/2 . . .let N2 be gas 1 and the unknown be gas 2. 79 mL/s / Rate gas 2 = (MM SO2 / MM N2)^1/2 = (64.1 g/mole / 28.0 g/mole)^1/2 = 1.51 Rate gas 2 = 79 mL/s / 1.51 = 52 mL/s
    N2 has an effusion rate of 79 mL/s. According to Graham's Law, Rate gas 1 / Rate gas 2 = (MM gas 2 / MM gas 1)^1/2 . . .let N2 be gas 1 and the unknown be gas 2. 79 mL/s / Rate gas 2 = (MM SO2 / MM N2)^1/2 = (64.1 g/mole / 28.0 g/mole)^1/2 = 1.51 Rate gas 2 = 79 mL/s / 1.51 = 52 mL/s
    3 answers · Chemistry · 9 months ago
  • What is the pH of a 0.216 M solution of hypobromous acid (HBrO)? Ka = 2.8 ✕ 10-9?

    For most weak acid problems, [H3O+] = sqrt (Ka x [HA]o) [H3O+] = sqrt ((2.8 x 10^-9)(0.216)) = 2.5 x 10^-5 M pH = -log [H3O+] = -log (2.5 x 10^-5) = 4.61
    For most weak acid problems, [H3O+] = sqrt (Ka x [HA]o) [H3O+] = sqrt ((2.8 x 10^-9)(0.216)) = 2.5 x 10^-5 M pH = -log [H3O+] = -log (2.5 x 10^-5) = 4.61
    1 answer · Chemistry · 9 months ago
  • Naming binary ionic compounds?

    Zirconium can exist in the +2 and +3 states although they are rare. In general, for any transition metal other than Zn, Cd, or Ag, which have only one oxidation state, use a Roman numeral.
    Zirconium can exist in the +2 and +3 states although they are rare. In general, for any transition metal other than Zn, Cd, or Ag, which have only one oxidation state, use a Roman numeral.
    2 answers · Chemistry · 1 year ago
  • If HCL gas is converted to Cl2 gas and H2 gas (@ constant temp & pressure) what will occur?

    The reaction is 2HCl(g) ==> Cl2(g) + H2(g) Note that there is no change in the number of gas moles ... 2 on each side. Thus, the piston would not move.
    The reaction is 2HCl(g) ==> Cl2(g) + H2(g) Note that there is no change in the number of gas moles ... 2 on each side. Thus, the piston would not move.
    1 answer · Chemistry · 1 year ago
  • How many mL of 1.63 M HNO3 will be needed to neutralize 191 mL of 3.55 M Sr(OH)2.?

    2HNO3(aq) + Sr(OH)2(aq) ==> Sr(NO3)2(aq) + 2H2O mmoles Sr(OH)2 = M Sr(OH)2 x mL Sr(OH)2 = (3.55)(191) = 678 mmoles Sr(OH)2 678 mmoles Sr(OH)2 x (2 mmoles HNO3 / 1 mmole Sr(OH)2) = 1360 mmoles HNO3 mmoles HNO3 = M HNO3 x mL HNO3 1360 = (1.63)(ml HNO3) mL HNO3 = 1360 / 1.63 = 832 mL
    2HNO3(aq) + Sr(OH)2(aq) ==> Sr(NO3)2(aq) + 2H2O mmoles Sr(OH)2 = M Sr(OH)2 x mL Sr(OH)2 = (3.55)(191) = 678 mmoles Sr(OH)2 678 mmoles Sr(OH)2 x (2 mmoles HNO3 / 1 mmole Sr(OH)2) = 1360 mmoles HNO3 mmoles HNO3 = M HNO3 x mL HNO3 1360 = (1.63)(ml HNO3) mL HNO3 = 1360 / 1.63 = 832 mL
    1 answer · Chemistry · 1 year ago
  • A balloon holds 32.7 kg of helium. What is the volume of the balloon if the pressure is 1.13 atm and the temperature is 23 °C?

    PV = nRT V = nRT / P n = moles of He = 32,700 g He x (1 mole He / 4.00 g He) = 818 moles He R = gas constant = 0.0821 L atm / K mole T = Kelvin temperature= 23 C + 273 = 300 K P = pressure in atm = 1.13 V = (818)(0.0821)(300) / 1.13 = 178,000 L
    PV = nRT V = nRT / P n = moles of He = 32,700 g He x (1 mole He / 4.00 g He) = 818 moles He R = gas constant = 0.0821 L atm / K mole T = Kelvin temperature= 23 C + 273 = 300 K P = pressure in atm = 1.13 V = (818)(0.0821)(300) / 1.13 = 178,000 L
    2 answers · Chemistry · 1 year ago
  • Calculate the volume of the gas when the pressure is 550 torr and the temperature is 40°C.?

    You also need to know how many moles of gas there are.
    You also need to know how many moles of gas there are.
    1 answer · Chemistry · 1 year ago
  • Chemistry lab question: Please Help! 1) After heating, the crucible is set on the lab bench where it is contaminated with the cleaning oil?

    The weight of the crucible is too high since it is contaminated with oil. If you then add the hydrated salt to the crucible and determine its mass by difference, that mass is OK. But if you burn off the oil during the dehydration step, then the apparent mass of anhydrous salt will be too low since both water and oil were burned off during the... show more
    The weight of the crucible is too high since it is contaminated with oil. If you then add the hydrated salt to the crucible and determine its mass by difference, that mass is OK. But if you burn off the oil during the dehydration step, then the apparent mass of anhydrous salt will be too low since both water and oil were burned off during the dehydration step.
    1 answer · Chemistry · 1 year ago
  • How do you calculate the amount of energy released by an electron returning to the ground state from an excited state in a copper atom?

    It depends on which excited state the electron is in. If you are referring to the intense green color that copper gives in a flame, its wavelength (lambda) is 5218 angstroms (5218 x 10^-10 m). nu (frequency) = c / lambda = (3.00 x 10^8 m/s) / (5218 x 10^-10 m) = 5.75 x 10^14 s^-1 The energy is found by E = (h)(nu) = (6.626 x 10^-34 J s)(5.75 x... show more
    It depends on which excited state the electron is in. If you are referring to the intense green color that copper gives in a flame, its wavelength (lambda) is 5218 angstroms (5218 x 10^-10 m). nu (frequency) = c / lambda = (3.00 x 10^8 m/s) / (5218 x 10^-10 m) = 5.75 x 10^14 s^-1 The energy is found by E = (h)(nu) = (6.626 x 10^-34 J s)(5.75 x 10^14 s^-1) = 3.6 x 10^-19 J
    1 answer · Chemistry · 1 year ago
  • It takes 5.2 minutes for a 1.000-g sample of 210Fr to decay to 0.350 g and follows a first order reaction. What is the half-life of 210Fr?

    Best answer: The integrated rate law for a first order decay is ln (Nt/No) = -kt, where Nt/No is the fraction of substance remaining at time t. ln (0.350 g / 1.000 g) = (-k)(5.2 min) -1.05 = -5.2k k = 0.20 min^-1 t1/2 = 0.693 / k = 0.693 / 0.20 min^-1 = 3.5 minutes
    Best answer: The integrated rate law for a first order decay is ln (Nt/No) = -kt, where Nt/No is the fraction of substance remaining at time t. ln (0.350 g / 1.000 g) = (-k)(5.2 min) -1.05 = -5.2k k = 0.20 min^-1 t1/2 = 0.693 / k = 0.693 / 0.20 min^-1 = 3.5 minutes
    1 answer · Chemistry · 1 year ago
  • Org. Chemistry: IUPAC naming?

    You are looking down the C1-C2 axis of a hydrocarbon. The carbon in back (C2) has methyl (CH3-) and isobutyl((CH3)2CH-CH2-) groups attached to it. If we redraw this with the first C as C1 (ignore the dots), we get .........CH3.........CH3 ..........|..............| CH3 - C - CH2 - C - H ..........|..............| .........H.............CH3 Notice... show more
    You are looking down the C1-C2 axis of a hydrocarbon. The carbon in back (C2) has methyl (CH3-) and isobutyl((CH3)2CH-CH2-) groups attached to it. If we redraw this with the first C as C1 (ignore the dots), we get .........CH3.........CH3 ..........|..............| CH3 - C - CH2 - C - H ..........|..............| .........H.............CH3 Notice that the longest carbon chain is 5 carbons long with CH3- groups on C2 and C4. This compound is 2,4-dimethylpentane. .........CH3.........CH3 ..........|..............| CH3 - C - CH2 - C - CH3 ..........|..............| .........H.............H
    1 answer · Chemistry · 1 year ago
  • (b) The volume of exactly 6 Skittles is 5.3 mL as measured in a 10.0 mL graduated cylinder.?

    150 Skittles x (5.3 mL / 6 Skittles) x (1 L / 1000 mL) = 0.13 L
    150 Skittles x (5.3 mL / 6 Skittles) x (1 L / 1000 mL) = 0.13 L
    2 answers · Chemistry · 1 year ago
  • How many moles of hydrogen atoms are in 4.63 grams of Ca(OH)2. The two is a subscript of the OH. ?

    Best answer: 4.63 g Ca(OH)2 x (1 mole Ca(OH)2 / 74.8 g Ca(OH)2) x (2 moles H / 1 mole Ca(OH)2) = 0.124 moles H
    Best answer: 4.63 g Ca(OH)2 x (1 mole Ca(OH)2 / 74.8 g Ca(OH)2) x (2 moles H / 1 mole Ca(OH)2) = 0.124 moles H
    1 answer · Chemistry · 1 year ago
  • How to make 1 L of 1 M sucrose?

    1 mole sucrose = 342 g sucrose So 1 M sucrose = 342 g sucrose / L Dissolve 342 g of sucrose in about 800 mL of water. Adjust the final volume to 1000 mL with water and mix well.
    1 mole sucrose = 342 g sucrose So 1 M sucrose = 342 g sucrose / L Dissolve 342 g of sucrose in about 800 mL of water. Adjust the final volume to 1000 mL with water and mix well.
    1 answer · Chemistry · 1 year ago
  • How to make 1 L of 0.1 sucrose?

    If you add 1 L of water to 342 g of sucrose, the final volume will be a lot higher than 1 L!!!! So disolve the 342 g in a lesser amount of water (500 to 800 ml), dissolve, and then add water until the final volume is 1000 mL.
    If you add 1 L of water to 342 g of sucrose, the final volume will be a lot higher than 1 L!!!! So disolve the 342 g in a lesser amount of water (500 to 800 ml), dissolve, and then add water until the final volume is 1000 mL.
    1 answer · Chemistry · 1 year ago
  • How to make 1 L of 0.1 M NaOH from 4 M NaOH? Is it 600 L of water and 400 L of NaOH? Sorry i'm confused on how to do the math. Thanks!?

    Dilution problems are readily solved using C1V1 = C2V2. C1 = concentration of concentrated solution = 4 M V1 = volume of concentrated solution = ? C2 = concentration of diluted solution = 0.1 M V2 = volume of diluted solution = 1 L C1V1 = C2V2 (4 M)(V1) = (0.1 M)(1 L) V1 = (0.1 M)(1 L) / (4 M) = 0.025 L = 25 mL Add 25 mL of 4 M NaOH to 975 mL of... show more
    Dilution problems are readily solved using C1V1 = C2V2. C1 = concentration of concentrated solution = 4 M V1 = volume of concentrated solution = ? C2 = concentration of diluted solution = 0.1 M V2 = volume of diluted solution = 1 L C1V1 = C2V2 (4 M)(V1) = (0.1 M)(1 L) V1 = (0.1 M)(1 L) / (4 M) = 0.025 L = 25 mL Add 25 mL of 4 M NaOH to 975 mL of water (total volume = 1000 mL = 1 L) and mix thoroughly.
    2 answers · Chemistry · 1 year ago
  • Chemistry atomic part cross word quiz?

    Best answer: Probably "isotopes".
    Best answer: Probably "isotopes".
    1 answer · Chemistry · 1 year ago