3. An industrial chemist introduces 2.0 atm H2 and 2.0 atm CO2 into a 1.00 L container at 25 ⁰C and then increases the temperature

Explanation:H₂ (g) + CO₂ (g) ⇌ H₂O (g) + CO (g)2 atm pressure at 25° or 298 K . If temperature increases to 700°C or 973K pressure = 2 x 973 / 298 = 6.53 atm H₂ (g) + CO₂ (g) ⇌ H₂O (g) + CO (g)6.53 - x    6.53 - x         x             x   x² / ( 6.53 - x )²   = .534 x  / ( 6.53 - x )  = .73 x = 4.77 - .73 x x = 2.76 pressure of hydrogen gas in equilibrium = 6.53 - 2.76 = 3.77 atm moles of hydrogen at this temperature and pressure n = PV / RT 3.77 x 1 / .082 x 973 = .04725 mole = .04725 x  2 gram = .0945 gram = 94.5 mili gram

0.0944 gram of H2Explanation:Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:2.0 atm x (973 K / 298 K) = 6.53 atmWhereKc = Kp because the moles of product equals the moles of reactants.At equilibriuim, the amounts areP(H2) = 6.53 - xP(CO2) = 6.53 - xP(H2O) = xP(CO) = xKc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]Take the square root of each side(.534)^0.5 = x / (6.53 - x) x = 0.731 (6.53 - x)x = 4.77 - 0.731x1.731x = 4.77x = 4.77 / 1.731 = 2.76 atmP(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atmP(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atmPV = nRTn = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H20.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g

1) moles of oxygen (n) = 1.86 molesExplanation:according to  Boyle's law the formula to solve this problem is:PV=nRtwhen P is the pressure which equal 1.25 atm and V is the volume which equal 37.5 Ln is the number of moles which we need to calculate itR is constant which equal 0.082t is the temperature in kelvin By substitution:1.25*37.5 = n * 0.082 * 307So n = 1.86 moles2) the volume of oxygen gas = 34 LExplanation:at standard temperature and pressure (STP) 1 mole of gas will equal = 22.4LSo when we have 1.5 moles of oxygen at standard temperature and pressure (STP) so we will estimate it like that 1.5 moles *22.4 L/ 1 mole = approximately 34 L 3) The volume of H2 = 2.29 L Explanation:according to the Balanced equation we can see that the molar ratio between Zn and H2 1 : 1so to know the number of moles of H2 we will get it for Zn first :number of moles Zn = mass of Zn / molar mass Zn                              = 5.98 / 65.39 =0.0914 molesso number of moles H2 = 0.09 molesby  substitution in the following formula:PV = nRT0.978 * V = 0.09 * 0.082 * 298so The volume of H2 = 2.29 L 4) Volume of O2 = 1.4 LExplanation:first we have to balance  the equation:2Na2O2 +2CO2 → 2Na2CO3 + O22 mole CO2 give 1 mole of O2 so the molar ratio is 2:1at STP 1 mole of gas will equal = 22.4 L              ??? moles of CO2 = 2.8n CO2 = 0.125 moles so n O2 = 0.125 /2 = 0.0625 molesso when 1 mole of as = 22.4               0.0625 moles O2 = ???Volume of O2 =0.0625 moles * 22.4 L/ 1 mole                         = 1.4 L5) the initial quantity of sodium metal used = 17.2 gramExplanation:at STP 1 mole of gas will equal = 22.4 Lso  moles of H2 equal ?? when 8.40 liters of H2 gas were producedso moles of H2  = 8.4/22.4 =0.375 molesand according to the balanced equation the molar ratio between H2 ans Na is 1 : 2 so number of moles for Na = 0.375 *2 = 0.75 molesto get the initial quantity of sodium metal (mass Na) = number of moles * molar mass mass Na = 0.75 moles * 23 gm/mole= 17.25 6) FalseExplanation:because STP means standard temperature and pressure. and the Standard temperature must be 273 K and the standard pressure must be 1 atm but in the question the temperature is 298 K not 273 K so , It is not a standard temperature7) 22.4 litersExplanation:1 mole of gas will equal = 22.4 Lbecause at the Standard temperature must be 273 K and the standard pressure must be 1 atmso V = nRT/P        =1 mole * 0.082 * 273 K / 1 aTm         = 22.4 liters

Not too sure about first question, Second is D. 41, 406 J.