• ### If two compounds are only "tied" together by a weak IMF (ex - London Dispersion), how low or high will their freezing points be compared to?

Best answer: Lower because with a weaker IMF it takes less energy to separate the molecules.
Best answer: Lower because with a weaker IMF it takes less energy to separate the molecules.
2 answers · Chemistry · 3 years 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 · 2 years ago
• ### Chemistry atomic part cross word quiz?

Best answer: Probably "isotopes".
Best answer: Probably "isotopes".
1 answer · Chemistry · 2 years 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 · 2 years ago
• ### Chemistry help please!?

Best answer: Write each one as a rate expression. delta [PH3] / dt . . . . .and . . . . .delta [H2] / dt Divide each by their coefficients in the balanced equation and equate the two. 1/4 delta [PH3] / dt = 1/6 delta [H2] / dt 1/4 (5.10 x 10^-4 M/s) = 1/6 delta [H2] / dt 1.28 x 10^-4 M/s = 1/6 delta [H2] / dt delta [H2] / dt = 6 x 1.28 x 10^-4 M/s... show more
Best answer: Write each one as a rate expression. delta [PH3] / dt . . . . .and . . . . .delta [H2] / dt Divide each by their coefficients in the balanced equation and equate the two. 1/4 delta [PH3] / dt = 1/6 delta [H2] / dt 1/4 (5.10 x 10^-4 M/s) = 1/6 delta [H2] / dt 1.28 x 10^-4 M/s = 1/6 delta [H2] / dt delta [H2] / dt = 6 x 1.28 x 10^-4 M/s = 7.65 x 10^-4 M/s
1 answer · Chemistry · 2 years ago
• ### How many seconds does it take the earth to make one orbit around the sun?

Best answer: 1. It takes the earth about 365.25 days to orbit the sun. 365.25 days x (24 hrs / 1 day) x (3600 s / 1 hr) = about 31,660,000 seconds 2. It takes the earth 24 hrs. to make one rotation. 24 hrs x (3600 s / 1 hr) = 86,400 seconds
Best answer: 1. It takes the earth about 365.25 days to orbit the sun. 365.25 days x (24 hrs / 1 day) x (3600 s / 1 hr) = about 31,660,000 seconds 2. It takes the earth 24 hrs. to make one rotation. 24 hrs x (3600 s / 1 hr) = 86,400 seconds
3 answers · Chemistry · 2 years ago
• ### The reaction is: Fe2O3 + 3C ---> 2Fe + 3CO If 22.5 moles of iron are produced in the reaction above, how many moles of carbon monoxide?

Best answer: 22.5 moles Fe x (3 moles CO / 2 moles Fe) = 33.8 moles CO produced
Best answer: 22.5 moles Fe x (3 moles CO / 2 moles Fe) = 33.8 moles CO produced
1 answer · Chemistry · 2 years ago
• ### Chemistry problem?

Best answer: D -- O2(g) only. The solid reactants PbS and C do not appear in the equilibrium constant expression.
Best answer: D -- O2(g) only. The solid reactants PbS and C do not appear in the equilibrium constant expression.
1 answer · Chemistry · 2 years ago
• ### What volume of 0.55 M KOH is needed to reach the endpoint with 0.665 g of HCN?

Best answer: 0.665 g HCN x (1 mole HCN / 31.0 g HCN) = 0.0215 moles HCN HCN reacts with KOH in a 1:1 mole ratio: HCN + KOH ==> H2O + KCN 0.0215 moles HCN x (1 mole KOH / 1 mole HCN) = 0.0215 moles KOH needed to titrate moles KOH = Molarity KOH x L KOH 0.0215 = 0.55 x L KOH L KOH = 0.0390 L KOH = 39 mL 0.55 M KOH
Best answer: 0.665 g HCN x (1 mole HCN / 31.0 g HCN) = 0.0215 moles HCN HCN reacts with KOH in a 1:1 mole ratio: HCN + KOH ==> H2O + KCN 0.0215 moles HCN x (1 mole KOH / 1 mole HCN) = 0.0215 moles KOH needed to titrate moles KOH = Molarity KOH x L KOH 0.0215 = 0.55 x L KOH L KOH = 0.0390 L KOH = 39 mL 0.55 M KOH
1 answer · Chemistry · 2 years ago
• ### Calculate the cell potential of the following voltaic (galvanic) cell. Zn2+(1.0 M)|Zn||Ni|Ni2+(.39 M)?

Best answer: If the cell is galvanic (E cell is positive), then Zn is the anode (oxidation) and Ni is the cathode (reduction). Zn(s) ==> Zn2+(aq) + 2e- Eo = +0.76 V Ni2+(aq) + 2e- ==> Ni(s) Eo = -0.23 V =================================== Zn(s) + Ni2+(aq) ==> Zn2+(aq) + Ni(s) . . .Eo cell = +0.53 V For nonstandard ion concentrations,... show more
Best answer: If the cell is galvanic (E cell is positive), then Zn is the anode (oxidation) and Ni is the cathode (reduction). Zn(s) ==> Zn2+(aq) + 2e- Eo = +0.76 V Ni2+(aq) + 2e- ==> Ni(s) Eo = -0.23 V =================================== Zn(s) + Ni2+(aq) ==> Zn2+(aq) + Ni(s) . . .Eo cell = +0.53 V For nonstandard ion concentrations, use the Nernst equation: E cell = Eo cell - 0.0592 / n log Q E cell = 0.53 V - 0.0592/2 log ([Zn2+] / [Ni2+]) = 0.53 - (0.0296)(log (1.0 / 0.39)) = 0.52 V
1 answer · Chemistry · 2 years ago
• ### Silver Ion Concentration?

Best answer: Think of the problem in two steps: (1) complete precipitation of Ag2CO3 because of the very small value of Ksp, and (2) dissolution of a very small amount of Ag2CO3 to satisfy Ksp. You correctly calculated that initially [Ag+] = 0.198 M and [CO3 2-] = 0.271 M. In a quantitative reaction, M . . . . . . .2Ag+ . . .+ . . .CO3 2- . .... show more
Best answer: Think of the problem in two steps: (1) complete precipitation of Ag2CO3 because of the very small value of Ksp, and (2) dissolution of a very small amount of Ag2CO3 to satisfy Ksp. You correctly calculated that initially [Ag+] = 0.198 M and [CO3 2-] = 0.271 M. In a quantitative reaction, M . . . . . . .2Ag+ . . .+ . . .CO3 2- . . .==> . . .Ag2CO3 Initial . . . .0.198 . . . . . . . .0.271 . . . . . . . . . . . . . . . . Change . .-0.198 . . . . . .-1/2(0.198) . . . . . . . . . . . . . . Final . . . . .^0 . . . . . . . . . .0.172 Then the precipitated Ag2CO3 very slightly redissolves: . . .M . . . . . .Ag2CO3(s) . . .==> . . .2Ag+ . . .+ . . .CO3 2- Initial . . . . . . . . . . . . . . . . . . . . . . . . .0 . . . . . . . . .0.172 Change . . . . . . . . . . . . . . . . . . . . . . .2x . . . . . . . . . .x Equilibrium . . . . . . . . . . . . . . . . . . . . 2x . . . . . . . .0.172+x Ksp = [Ag+]^2[CO3 2-] = (2x)^2(0.172+x) = Ksp = 8.1 x 10^-12 Since the +x term is negligible compared to 0.172, (4x^2)(0.172) = 8.1 x 10^-12 0.688x^2 = 8.1 x 10^-12 x^2 = 1.2 x 10^-11 x = 3.5 x 10^-6 [Ag+] = 2x = 7.0 x 10^-6 M
1 answer · Chemistry · 2 years ago
• ### Name the Ketone? Chemistry?

1 answer · Chemistry · 2 years ago
• ### How many unbonded pairs of electrons are in the Lewis structure for potassium tetraflu- oroborate?

Best answer: 12 since there are 3 unbonded pairs on each F in BF4-.
Best answer: 12 since there are 3 unbonded pairs on each F in BF4-.
2 answers · Chemistry · 3 years ago
• ### HELPPPP ME PLEASEEEEE !!!!!!! HELP ASAP!! :) n2(g)+3h2(g)→2nh3(g)?

Best answer: Notice that it takes 3 moles of H2 to react with 1 mole of N2. 0.220 moles N2 x (3 moles H2 / 1 mole N2) = 0.660 moles H2 needed to completely react with N2. We have more than enough H2 (0.725 moles vs. 0.660 moles needed). Therefore, N2 is the limiting reactant that decided how much product we can make and we will have some H2 left... show more
Best answer: Notice that it takes 3 moles of H2 to react with 1 mole of N2. 0.220 moles N2 x (3 moles H2 / 1 mole N2) = 0.660 moles H2 needed to completely react with N2. We have more than enough H2 (0.725 moles vs. 0.660 moles needed). Therefore, N2 is the limiting reactant that decided how much product we can make and we will have some H2 left over at the end of the reaction. 0.220 moles N2 x (2 moles NH3 / 1 mole N2) = 0.440 moles NH3 produced Excess moles H2 = Total moles H2 - moles of H2 needed to react = 0.725 - 0.660 = 0.065 moles
1 answer · Chemistry · 3 years ago
• ### You mix 26.83 g of calcium and 56.9 g of iodine to form calcium iodide. How many moles of calcium iodide will be produced?

Best answer: 26.83 g Ca x (1 mole Ca / 40.08 g Ca) = 0.6694 moles Ca 56.9 g I2 x (1 mole I2 / 253.8 g I2) = 0.2242 moles I2 The reaction is: Ca + I2 ==> CaI2 . . .note that Ca and I2 react in a 1:1 mole ratio. Based on the number of moles calculated above, obviously we will run out of I2 first. 0.2242 moles I2 x (1 mole CaI2 / 1 mole I2) =... show more
Best answer: 26.83 g Ca x (1 mole Ca / 40.08 g Ca) = 0.6694 moles Ca 56.9 g I2 x (1 mole I2 / 253.8 g I2) = 0.2242 moles I2 The reaction is: Ca + I2 ==> CaI2 . . .note that Ca and I2 react in a 1:1 mole ratio. Based on the number of moles calculated above, obviously we will run out of I2 first. 0.2242 moles I2 x (1 mole CaI2 / 1 mole I2) = 0.2242 moles CaI2 formed
1 answer · Chemistry · 2 years ago
• ### How many grams of CO2 can 500.0 g of LiOH absorb?

Best answer: LiOH + CO2 ==> LiHCO3 500.0 g LiOH x (1 mole LiOH / 23.95 g LiOH) x (1 mole CO2 / 1 moles LiOH) x (44.01 g CO2 / 1 mole CO2) = 918.8 g CO2
Best answer: LiOH + CO2 ==> LiHCO3 500.0 g LiOH x (1 mole LiOH / 23.95 g LiOH) x (1 mole CO2 / 1 moles LiOH) x (44.01 g CO2 / 1 mole CO2) = 918.8 g CO2
3 answers · Chemistry · 2 years ago
• ### Which of the following potential solutes will dissolve into water?

Best answer: Be3N2 slowly decomposes in water to form Be(OH)2(aq). Propanol dissolves in water but butane will not. Be(OH)2 is only slightly soluble in water so its solution would be considered as a weak electrolyte.
Best answer: Be3N2 slowly decomposes in water to form Be(OH)2(aq). Propanol dissolves in water but butane will not. Be(OH)2 is only slightly soluble in water so its solution would be considered as a weak electrolyte.
1 answer · Chemistry · 2 years ago
• ### If i have a 52 liter container that contains 48 moles of gas at a temperature of 150C, what is the pressure inside the container?

Best answer: Use the ideal gas law: PV = nRT. P = gas pressure in atm n = moles of gas = 48 V = gas volume in L = 52 T = Kelvin temperature = 150 C + 273 = 423 R = universal gas constant (0.0821 L atm / K mole) P = nRT / V = (48)(0.0821)(423) / 52 = 32 atm
Best answer: Use the ideal gas law: PV = nRT. P = gas pressure in atm n = moles of gas = 48 V = gas volume in L = 52 T = Kelvin temperature = 150 C + 273 = 423 R = universal gas constant (0.0821 L atm / K mole) P = nRT / V = (48)(0.0821)(423) / 52 = 32 atm
1 answer · Chemistry · 2 years ago
• ### If 4.66 grams of oxygen gas is reacted with carbon monoxide,what volume of carbon dioxide is produced at 36 c at 56 kpa?

Best answer: Write the balanced equation: 2CO + O2 ==> 2CO2 4.66 g O2 x (1 mole O2 / 32.0 g O2) = 0.146 moles O2 0.146 moles O2 x (2 moles CO2 / 1 mole O2) = 0.291 moles CO2 Use the ideal gas law to calculate the volume. P = 56 kPa x (1 atm / 101.3 kPa) = 0.55 atm T = 36 C (+ 273) = 309 K PV = nRT V = nRT / P = (0.291 moles)(0.0821 L atm / K... show more
Best answer: Write the balanced equation: 2CO + O2 ==> 2CO2 4.66 g O2 x (1 mole O2 / 32.0 g O2) = 0.146 moles O2 0.146 moles O2 x (2 moles CO2 / 1 mole O2) = 0.291 moles CO2 Use the ideal gas law to calculate the volume. P = 56 kPa x (1 atm / 101.3 kPa) = 0.55 atm T = 36 C (+ 273) = 309 K PV = nRT V = nRT / P = (0.291 moles)(0.0821 L atm / K mole)(309 K) / (0.55 atm) = 13 L
1 answer · Chemistry · 2 years ago
• ### Plz help.?

Best answer: If you add acid (HCl) to a buffer, it will react with the buffer base, in or case, acetate ion. The initial [HAc] = 0.100 moles / 0.100 L = 1.00 M. The initial [Ac-] = 0.110 moles / 0.100 L = 1.10 M initial mmoles HAc = M HAc x mL HAc = (1.00)(100 mL) = 100 mmoles HAc initial mmoles Ac- = M Ac- x mL Ac- = (1.10)(100 mL) = 110 mmoles... show more
Best answer: If you add acid (HCl) to a buffer, it will react with the buffer base, in or case, acetate ion. The initial [HAc] = 0.100 moles / 0.100 L = 1.00 M. The initial [Ac-] = 0.110 moles / 0.100 L = 1.10 M initial mmoles HAc = M HAc x mL HAc = (1.00)(100 mL) = 100 mmoles HAc initial mmoles Ac- = M Ac- x mL Ac- = (1.10)(100 mL) = 110 mmoles Ac- mmoles HCl to be added = M HCl x mL HCl = (0.300)(40.00) = 12.0 mmoles HCl Here's the reaction: mmoles . . . .H+ . . .+ . . .Ac- . . .==> . . .HAc initial . . . . . 12 . . . . . . . 110 . . . . . . . . .100 change . . . .-12 . . . . . . .-12 . . . . . . . . .+12 final . . . . . . .0 . . . . . . . .98 . . . . . . . . . .112 The pH of any buffer is readily found using the Henderson-Hasselbalch equation: pH = pKa + log (amt. buffer base / amt. buffer acid) Ka for HAc = 1.8 x 10^-5; pKa = -log Ka = 4.74 pH = 4.74 + log (mmoles Ac- / mmoles HAc) = 4.74 + log (98 / 110) = 4.69 Before adding the acid, the pH was 4.74 + log (110 / 100) = 4.78, so the pH changed by only 0.09 units. If you added 12 mmoles of HCl (in 40 mL of solution) to 120 mL of water, [H+] = 12 mmoles / 160 mL = 0.075 M; pH = -log [H+] = -log (0.075) = 1.1!
1 answer · Chemistry · 2 years ago