User:Toddlertoddy/Sandbox

$$(1.9\ \cancel{angstrom})\left (\frac{10^{-10}\ \cancel{m}}{1\ \cancel{angstrom}}\right)\left (\frac{10^{9}\ nm}{1\ \cancel{m}}\right) = 1.9 \times 10^{-1}\ nm$$ $$(1.9\ \cancel{angstrom})\left (\frac{10^{-10}\ \cancel{m}}{1\ \cancel{angstrom}}\right)\left (\frac{10^{12}\ pm}{1\ \cancel{m}}\right) = 1.9 \times 10^2\ pm$$ $$Kr = (1.0\ \cancel{mm})\left (\frac{10^{-3}\ \cancel{m}}{1\ \cancel{mm}}\right)\left (\frac{10^{10}\ \cancel{angstrom}}{1\ \cancel{m}}\right)\left (\frac{0.5263157895\ Kr\ atom}{1\ \cancel{angstrom}}\right)=5.3 \times 10^6\ Kr\ atoms$$ $$\frac{1.609344\ km}{1\ mi}$$ $$\frac{0.946352946\ L}{1\ qt}$$ $$0.076\ L = (0.076\ \cancel{L})\left (\frac{1000\ mL}{1\ \cancel{L}}\right) = 76\ mL$$ $$5.0 \times 10^{-8}\ m = (5.0 \times 10^{-8}\ \cancel{m})\left (\frac {10^{9}\ nm}{1\ \cancel{m}}\right) = 5.0 \times 10^{1}\ nm$$ $$6.88 \times 10^5\ ns = (6.88 \times 10^5\ \cancel{ns})\left (\frac {10^{-9}\ s}{1\ \cancel{ns}}\right) = 6.88 \times 10^{-4}\ s$$ $$0.50\ lb = (0.50\ \cancel{lb})\left (\frac{453.59237\ g}{1\ \cancel{lb}}\right) = 2.3 \times 10^2\ g $$ $$1.55\ kg/m^3 = \left (\frac {1.55\ \cancel{kg}}{\cancel{m^3}}\right)\left (\frac{1000\ g}{1\ \cancel{kg}}\right)\left (\frac{0.001\ \cancel{m^3}}{1\ L}\right) = 1.55\ g/L$$ $$5.850\ gal/hr = \left (\frac{5.850\ \cancel{gal}}{\cancel{hr}}\right)\left (\frac{3.78541178\ L}{1 \cancel{gal}}\right)\left (\frac{0.000277777778\ \cancel{hr}}{1\ s}\right) = 6.151 \times 10^{-3}\ L/s$$ $$1.19\ g/L \times (12.5\ ft \times 15.5\ ft \times 8.0\ ft)= \left (\frac{1.19\ \cancel{g}}{\cancel {L}}\right)(1600\ \cancel{ft^3})\left (\frac{28.3168466\ \cancel{L}}{1\ \cancel{ft^3}}\right)\left (\frac{0.001\ kg}{1 \cancel{g}}\right) = 54\ kg$$ $$5.00\ days = (5.00\ \cancel{days})\left (\frac{86 400\ s}{1\ \cancel{day}}\right) = 4.32 \times 10^5\ s$$ $$0.0550\ mi = (0.0550\ \cancel{mi})\left (\frac{1609.344\ m}{1 \cancel{mi}}\right) = 88.5\ m$$ $$$1.89/gal = \left (\frac{$1.89}{\cancel {gal}}\right)\left (\frac{0.264172052\ \cancel{gal}}{1\ L}\right) = $0.50/L $$ $$0.510\ in/ms = \left (\frac {0.510\ \cancel{in}}{\cancel{ms}}\right)\left (\frac{2.54 \times 10^{-5}\ km}{1 \cancel{in}}\right)\left (\frac{3 600 000\ \cancel{ms}}{1\ hr}\right) = 46.6\ km/hr$$ $$22.50\ gal/min = \left (\frac{22.50\ \cancel{gal}}{\cancel {min}}\right)\left (\frac{3.78541178\ L}{1\ \cancel{gal}}\right)\left (\frac{0.016667\ \cancel{min}}{1\ s}\right) = 1.420\ L/s$$ $$0.02500\ ft^3 = (0.02500\ \cancel{ft^3})\left (\frac{28 316.8466\ cm^3}{1\ \cancel{ft^3}}\right) = 707.9\ cm^3$$ $$31\ gal = (31\ \cancel{gal})\left (\frac{3.78541178\ L}{1\ \cancel{gal}}\right) = 1.2 \times 10^2\ L$$ $$6\ mg/kg \times 150\ lb = \left (\frac{6\ mg}{\cancel{kg}}\right)\left (\frac{0.45359237\ \cancel{kg}}{1\ \cancel{lb}}\right)(150\ \cancel{lb}) = 4 \times 10^2\ mg$$ $$\frac{254\ mi}{11.2\ gal} =\left (\frac{254\ \cancel{mi}}{11.2\ \cancel{gal}}\right)\left(\frac{1.609344\ km}{1\ \cancel{mi}}\right)\left(\frac{0.264172052\ \cancel{gal}}{1\ L}\right) = 9.64\ km/L$$ $$50\ cups/lb \times 1\ g= \left (\frac{50\ \cancel {cups}}{\cancel {lb}}\right)\left (\frac{236.588237\ mL}{1\ \cancel{cup}}\right)\left (\frac{0.00220462262\ \cancel{lb}}{1\ \cancel{g}}\right)(1\ \cancel g) = 26\ mL$$ $$(1.9\ \cancel{angstrom})\left (\frac{0.1\ nm}{1\ \cancel{angstrom}}\right)= 0.19\ nm$$ $$(1.9\ \cancel{angstrom})\left (\frac{100\ pm}{1\ \cancel{angstrom}}\right) = 1.9 \times 10^2\ pm$$ $$Kr = \frac{1.0\ \cancel{mm}}{(3.8\ \cancel{angstrom})\left (\frac{10^{-7}\ \cancel{mm}}{1\ \cancel{angstrom}}\right)} \times 1\ Kr\ atom =2.6 \times 10^6\ Kr\ atoms$$ $$V = \frac{4\pi r^3}{3} = \frac{4\pi}{3} \times \left((1.9\ \cancel{angstrom})\left (\frac{10^{-8}\ cm}{1\ \cancel{angstrom}}\right)\right)^3 = 2.9 \times 10^{-23}\ cm^3$$ $$ratio\ of\ ZnSO_4\ to\ H_2O = \frac {4.52\ g\ of\ ZnSO_4 \left(\frac {1\ mol\ of\ ZnSO_4}{161.5\ g\ of\ ZnSO_4}\right) }{0.56\ g\ of\ H_2O \left(\frac {1\ mol\ of\ H_2O}{18.0\ g\ of\ H_2O}\right)} = \frac{1.00\ mol\ of\ ZnSO_4}{1.1\ mol\ of\ H_2O} \approx \frac{1\ mol\ of\ ZnSO_4}{1\ mol\ of\ H_2O}$$<br\> $$k_{oxygen} = \frac{C_{oxygen}}{P_{oxygen}} = \frac{0.0430\ g/L}{760\ torr} = 5.6579 \times 10^{-5}\ \frac{g/L}{torr}$$<br\> $$C_{oxygen} = \left(5.6579 \times 10^{-5}\ \frac{g/L}{torr}\right)\left(156\ torr\right) = 0.00883\ g/L$$<br\> $$k_{nitrogen} = \frac{C_{nitrogen}}{P_{nitrogen}} = \frac{0.0190\ g/L}{760\ torr} = 2.50 \times 10^{-5}\ \frac{g/L}{torr}$$<br\> $$C_{nitrogen} = \left(2.50 \times 10^{-5}\ \frac{g/L}{torr}\right)\left(586\ torr\right) = 0.0146\ g/L$$<br\> $$amount\ of\ O_2 = (0.00883\ g/L)(0.100\ L) = 0.000883\ g\ dissolved$$<br\> $$amount\ of\ N_2 = (0.0146\ g/L)(0.100\ L) = 0.00146\ g\ dissolved$$<br\> $$C_{oxygen\ in\ air} = C_{pure}\ \times\ ratio\ of\ pressures = (0.0430\ g/L)\left (\frac{156\ torr}{760\ torr}\right) = 0.00883\ g/L$$<br\> $$C_{nitrogen\ in\ air} = C_{pure}\ \times\ ratio\ of\ pressures = (0.0190\ g/L)\left (\frac{586\ torr}{760\ torr}\right) = 0.0146\ g/L$$<br\> $$b.p. = 100.0^{\circ} C + \left(0.51 \frac{^{\circ}C\ kg\ solvent}{mol\ solute}\right)\left(\frac{16\ mol\ solute}{kg\ solvent}\right) = 108.2^{\circ} C$$<br\> $$\Delta t = 84.30^{\circ} C - 61.20^{\circ} C = 23.10^{\circ} C$$<br\> $$m = \frac{\Delta t}{k_b} = \frac{23.10^{\circ} C}{\frac{3.63^\circ C\ kg\ solvent}{mol\ solute}} = \frac{6.36\ mol\ solute}{kg\ solvent}$$<br\> $$molar\ mass\ of\ solute = \frac{11\ g}{(0.100 kg)\left(\frac{6.36\ mol\ solute}{kg\ solvent}\right)} = 17\ g/mol$$ $$f.p. = 0.0^{\circ} C - \left(1.86 \frac{^{\circ}C\ kg\ solvent}{mol\ solute}\right)\left(\frac{16\ mol\ solute}{kg\ solvent}\right) = -30.^{\circ} C$$<br\> $$\Delta t = 5.45^{\circ} C - 4.13^{\circ} C = 1.32^{\circ} C$$<br\> $$m = \frac{\Delta t}{k_f} = \frac{1.32^{\circ} C}{\frac{5.07^\circ C\ kg\ solvent}{mol\ solute}} = \frac{0.260\ mol\ solute}{kg\ solvent}$$<br\> $$molar\ mass\ of\ solute = \frac{3.46\ g}{(0.0850\ kg)\left(\frac{0.260\ mol\ solute}{kg\ solvent}\right)} = 157\ g/mol$$ $$f.p. = 178.4^{\circ} C - \left(37.7 \frac{^{\circ}C\ kg\ solvent}{mol\ solute}\right)\left(\frac{0.260\ mol\ solute}{kg\ solvent}\right) = 168.6^{\circ} C$$<br\> $$PV = nRT \to \frac{n}{V} = \frac{P}{RT}$$<br\> $$\frac{n}{V} = \frac{(26\ torr)\left(\frac{1\ atm}{760\ torr}\right)}{(0.08206\ L\ atm/K\ mol)(293\ K)} = 0.0014\ M$$<br\> $$vapour\ pressure\ = (23.8\ torr)\left(\frac{(1.00\ \times\ 10^3\ g)\left(\frac{1\ mol}{18.02\ g}\right)}{(1.00\ \times\ 10^3\ g)\left(\frac{1\ mol}{18.02\ g}\right) + (18.0\ g)\left(\frac{1\ mol}{180\ g}\right)}\right) = 23.8\ torr$$<br\> $$b.p. = 100.0^{\circ} C + \left(0.51 \frac{^{\circ}C\ kg\ solvent}{mol\ solute}\right)\left(\frac{2.00\ mol\ solute}{kg\ solvent}\right) = 101.0^{\circ} C$$<br\> $$f.p. = 0.00^{\circ} C - \left(1.86 \frac{^{\circ}C\ kg\ solvent}{mol\ solute}\right)\left(\frac{2.00\ mol\ solute}{kg\ solvent}\right) = -3.72^{\circ} C$$<br\> $$\Delta t = 5.45^{\circ} C - 3.45^{\circ} C = 2.00^{\circ} C$$<br\> $$m = \frac{\Delta t}{k_f} = \frac{2.00^{\circ} C}{\frac{1.86^\circ C\ kg\ solvent}{mol\ solute}} = \frac{1.07527\ mol\ solute}{kg\ solvent}$$<br\> $$molar\ mass\ of\ solute = \frac{12.00\ g}{(0.2000 kg)\left(\frac{1.07527\ mol\ solute}{kg\ solvent}\right)} = 55.8\ g/mol$$ $$\Pi = (1)(0.010\ M)(0.08206\ L\ atm/K\ mol)(298\ K) = (0.24\ atm)\left(\frac{760\ torr}{1\ atm}\right) = 1.8 \times 10^2\ torr$$<br\> $$\Delta t = ik_fm \to 0.261 = i\left(\frac{1.86^\circ C\ kg\ solvent}{mol\ solute}\right)\left(\frac{0.125\ mol\ solute}{kg\ solvent}\right)\to i = 1.12$$<br\> $$336.0\ torr = (400.0\ torr)X \to X = 0.8400$$<br\> $$0.1600 = \frac{mol\ of\ solute}{mol\ of\ solute + (34.88\ g)\left(\frac{1\ mol}{108.97\ g}\right)} \to mol\ of\ solute = 0.06095 mol$$<br\> molar\ mass = \frac{19.35\ g}{0.06095\ mol} = 317.5 g/mol