exam 3
If 6.13 mol of a gas has a volume of 48.7 L and a pressure of 1.85 atm, what is the temperature?Use R = 0.0821 atm • L/mol • K.
( 1.85 atm*48.7 L)/(6.13 mol*0.0821 atm • L/mol • K.) =179.02
Hydrogen holds promise as an environment friendly fuel. How many grams of H2 gas are present in a 42.0 L fuel tank at a pressure of 2861 lb/in2 (psi) at 20.0°C? Assume that 1 atm = 14.7 psi.
((2861/ 14.7 )atm*42.0 L *2.016gmolh2)/(0.0821aatm*mol^-1*k^-1*293.15k) =686g
A sample of gas has a volume of 1.60 L under 2.35 atm pressure at 45°C. If the gas is then expanded to occupy a volume of 12.2 L at 0.515 atm, what is the temperature in degrees Celsius? Assume the number of moles of gas is held constant.
(0.515 atm*12.2 L*(45+273.15))/(1.60 L*2.35 atm ) 531.63-273=258.63c
A balloon is filled to a volume of 13.11 L at a temperature of 27.1ºC. If the pressure in the balloon is measured to be 2.200 atm, how many moles of gas are contained inside the balloon?
(2.200 atm*13.11 L )/(0.0821atm*l/mol*k*300.25) 1.17mole
Use the following data to determine the ΔΗ° for the conversion of diamond into graphite: C(s,diamond)+O2(g)CO2(g) ΔΗ° = -395.4 kJ 2CO2(g)2CO(g)+O2(g) ΔΗ° = 566.0 kJ 2CO(g)C(s,graphite)+CO2(g) ΔΗ° = -172.5 kJ $$C(s,diamond)C(s,graphite) ΔΗ° = kJ
-1.90kj
Calculate ΔH°rxn for the reaction $$2Ni(s)+2S(s)+3O2(g)2NiSO3(s) from the following information: 1. NiSO3(s)NiO(s)+SO2(g) ΔHorxn=156 kJ 2. S(s)+O2(g)SO2(g) ΔHorxn=−297 kJ 3. Ni(s)+12O2(g)NiO(s) $$ΔHorxn=−241 kJ ΔH°rxn =
-1390
A gaseous mixture consists of 1.21 mol of helium, 0.36 mol of neon, and 1.58 mol of argon. What is the mole fraction of helium in the mixture?
0.384
Convert the following pressures into atmospheres. 723.0 mmHg
0.9513atm
Convert each of the following physical properties to the appropriate units for use in the ideal gas law equation. 755 mmHg =
0.993atm
The diagram below shows how a chemical reaction in a cylinder with a piston that affects the volume of the system. Use this diagram to answer the following questions. Which of the following statements best describes the reaction?
1)=Work is done on the system by the surroundings. The equation used to calculate pressure-volume work is w=−PΔV=−Δ.
The complete combustion of 1.15 g of cinnamaldehyde (C9H8O, one of the compounds in cinnamon) in a bomb calorimeter (Ccalorimeter = 3.95 kJ/°C) produced an increase in temperature of 10.4 °C. Calculate the molar enthalpy of combustion of cinnamaldehyde (ΔHcomb) (in kilojoules per mole of cinnamaldehyde).
1.15/132.16=0.008701574mol 3.95kj/c*10.4c =41.08ks/ 0.00870157=-4720.98kj/mol
At fixed temperature, how much faster does NO effuse than NO2?
1.24 times faster
The addition of 115.0 g potassium fluoride to 597.0 mL of water (d = 1.00 g/mL) causes the temperature to rise 3.31 °C. What is the ΔHrxn for KF? (Assume cs = 4.18 J/g oC.)
115.0g/58.097=1.9794mol 115.0+597.0=712x4.18x3.31=9851.0896j =-4976.8j*1kt/1000j -4.97kt/mol -5.0kt/mol
An insulated container is used to hold 43.1 g of water at 16.1 °C. A sample of copper weighing 17.4 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.0°C. The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 0.385 J/(g·°C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.
17.4g*0.385j(t+100.0)=43.1g*4.185j*(t-16.1c) 43.1g-17.4g=add 4.185j-0.385j=add 27.31*(t-16.1c) =-439.691c =27.31t t+100.0c= (t-16.1c) 439.691c+100.0c=t+27.31t 539.691c/28.31=28.31c/28.31 19.06c=t
19.0 J of heat is added to 10.00 g of gold, which is initially at 25°C.The heat capacity of gold is 25.41 J/(mol • °C). What would be the final temperature of the gold?
19.0j/(0.0508molau*25.41*j/mol*c)=14.7c 14.7c+25c=39.7c
The diagram below shows how a chemical reaction in a cylinder with a piston that affects the volume of the system. Use this diagram to answer the following questions. If the reaction is endothermic, what happens to the internal energy of the system when the reaction is proceeding?
2)=It increases.
A flashbulb of volume 1.80 mL contains O2(g) at a pressure of 2.30 atm and a temperature of 30.0 °C. How many grams of O2(g) does the flashbulb contain? (do in scn
2.30 atm *1.80x10-3 L/(0.0821L*atm/mol*k*303.15k) =1.66*10^-4mol mass of o2=32 32g/mol*1.66*10^-4mol =5.32*10^-3g o2
A chamber initially at 0.884 atm and 22.4 L of He undergoes a change in pressure such that the final volume is 49.2 L. What is the final pressure assuming that temperature and the number of moles are constant?
2⋅ From Boyles's law, P1V1=P2V2 P1=0.884 atm⋅P2=? V1=22.4L, V2=49.2 LL P2=0.884atmx 22.4/ 49.2 L p2=0.402atm
Consider a neutralization reaction in which NaOH(aq) is mixed with H2SO4(aq) in a coffee-cup calorimeter. If 159.0 g of liquid in the coffee-cup calorimeter rises 8.50 °C, how much heat is released by the neutralization reaction?
4.185j/gc 8.50c 159.0g 159.0x4.184*8.50=5654.676j*1kj/1000j =5.65kj
A 58.5 g block of an unknown metal is heated in a hot water bath to 100.0°C. When the block is placed in an insulated vessel containing 130.0 g of water at 25.0°C, the final temperature is 28.0°C. Determine the specific heat of the unknown metal. The cs for water is 4.18 J/g°C.
58.5g(c1)(100.0-28.0)c=(130.0g)(4.18j/gc)(28..0-25.0)c 58.5g(c1)72.0c=(130.0g)(4.18j/gc)3c (543.4j/gcx3)/(58.5gx72.0c) 0.387j/gc
Calculate ΔE for the following situations: q = 6.400 kJ; w = 0.8000 J
6.400kf*1000or10^3j/1kj+0.8000j=6400.8j
Calculate the root-mean-square speed of helium at 331 K in meters per second, and compare your result with the root-mean-square speed of nitrogen at the same temperature.
</3(8.314*331)/(4*10^-3) =1430m/s
The air pressure in the tires of an automobile is adjusted to 27.0 psi at a gas station in San Diego, where the air temperature is 68°F. The air in the tires is at the same temperature as the atmosphere. The automobile is then driven east along a hot desert highway, and the temperature inside the tires reaches 160 °F. What is the pressure in the tires? (To calculate the pressure of a tire at a new temperature, we use Amontons's law in the form:P1/T1=P2/T2)
= 27.0 psi x ( 1 atm / 14.6959 psi) p1=1.8372 Now, Converting the temperature from oF to K, T1 = (68 -32) /1.8 = 20 oC + 273.15 = 293.15 K T2 = ( 160 - 32) *(5/9) = 71.11 oC + 273.15 = 344.26k Now, We know the Gay-Lussac's law formula for P and T, P1 /T1 = P2 /T2 Rearranging the formula, P2 = (P1T2) / T1 P2 = (1.8372 atm x = 344.26k) / 293.15 K P2 = 2.1575 atm Converting atm to psi, = 2.1575 atm x (14.6959 psi/ 1 atm) = 31.71 psi
Convert the following pressures into atmospheres. 8.10 kPa
=0.0799atm
Convert each of the following physical properties to the appropriate units for use in the ideal gas law equation. 267.5 mL =
=0.2675l
How much heat (in joules) is needed to increase the temperature of a 2.75 mol sample of solid iron from 24.20°C to 43.00°C? The specific heat capacity of solid iron is 0.444 J/g • K.
=1290j
Convert each of the following physical properties to the appropriate units for use in the ideal gas law equation. 100°C =
=373.15k
Solid potassium hydroxide may be used to unclog sink drains. The dissolution of potassium hydroxide in water is quite exothermic. What are the signs of ΔHsys and qsurr? ΔHsys
=negative
Solid potassium hydroxide may be used to unclog sink drains. The dissolution of potassium hydroxide in water is quite exothermic. What are the signs of ΔHsys and qsurr? qsurr
=positive
The behavior of a real gases deviates from the behavior of an ideal gas. Two places where this behavior difference is most evident are:1) when the total volume is increasingly made up of volume occupied by molecules.2) when the intermolecular forces between molecules becomes more significant to molecule movement.
AnalyzePV/RT is a rearrangement of the ideal gas equation and would be equal to n. Since the number of moles of gas is not changing with pressure, it should be apparent that the ideal gas equation yields insufficient results for real gases. Working within our ideally constructed equation, we must determine how the predicted number of ideal gas molecules differs when our molecules exhibit significant real properties. SolveFor an ideal gas, PV/RT should be constant with respect to pressure.Assuming our real gases have a similar type of intermolecular force, their masses should determine the relative intermolecular force strength. Stronger intermolecular forces yield smaller pressures, which decrease the magnitude of PV/RT.As the free volume-to-total-volume ratio decreases, the magnitude of PV/RT increases. This occurs for all real gases as pressure increases.Think About ItOnly molecules at low pressure with weak intermolecular forces are described adequately by the ideal gas equation.
Which of the drawings in the figure best illustrates what happens when the temperature of a helium-filled rubber balloon is increased at constant pressure?
Answer: According to Charles's gas law (V₁/T₁ = V₂/T₂), if the pressure is held constant, the volume and temperature of a given mass of gas increase proportionally. So the answer is B. The molecules get farther apart (ballon is larger) Collect and OrganizeWhen the temperature outside of the balloon increases, does the balloon (a) shrink while keeping the molecules randomly distributed in the balloon, (b) expand, or (c) shrink and condense the molecules of gas into liquid? Analyze Charles's law states that volume and temperature of a gas are directly proportional. Solve If we increase the temperature, we will increase the volume of the balloon by Charles's law. This is represented by (b) in the diagram. Think About It If the temperature is increased too much, the balloon will burst.
A balloon is partly inflated with 4.95 liters of helium at sea level where the atmospheric pressure is 1010 mbar. The balloon ascends to an altitude of 3.00 × 103 meters, where the pressure is 863 mbar. What is the volume of the helium in the balloon at the higher altitude? Assume that the temperature of the gas in the balloon does not change in the ascent.
Apply ideal gas law ratios V1P1 = V2P2 4.95*1010 = V2*863 V2 = 1010mbar* 4.95l/863mbar) = 5.79316 L
dentify the correct relationship between pressure and volume. Temperature and the number of moles are constant.
Boyle's Law it is defines that at a fixed temperature the volume of gas is inversely proportional to the pressure exerted by the gas Explanation: Boyle's law tells that the pressure decreases as the volume of gas increases. According to Boyles's law, pressure and volume are related as inversely proportional at constant temperature and moles. So the given graph Line 2 is correct.
The following reactions are carried out at atmospheric pressure (external) inside the chamber of a piston. Will the piston plunger have moved out (expanded) or moved in (contracted) after the reaction has occurred and the system was allowed to return to its initial temperature of 110°C? .
C(s)+O2(g)CO2(g) The piston is unchanged.
The following reactions are carried out at atmospheric pressure (external) inside the chamber of a piston. Will the piston plunger have moved out (expanded) or moved in (contracted) after the reaction has occurred and the system was allowed to return to its initial temperature of 110°C?,
C2H5OH(g)+3O2(g)2CO2(g)+3H2O(g) .B. The piston expanded.
Radon is a naturally occurring radioactive gas found in the ground and in building materials. It is easily inhaled and emits alpha particles when it decays. Cumulative radon exposure is a significant risk factor for lung cancer.
Calculate the density of radon at 323 K and 1.05 atm of pressure. =8.79 Are radon concentrations likely to be greater in the basement or on the top floor of a building? =basement
Consider the following diagram containing a mixture of two gases, and answer the following questions.
Calculate the partial pressure of red spheres in Cylinder A. The internal pressure is 0.81 atm. number of Xe = 8 number of Kr = 4 mole fractio of Xe = 8 / 8 + 4 = 0.67 partial pressure of Xe = mole fraction x total pressure = 0.67 x 0.81 = 0.5427atm Calculate the partial pressure of blue spheres in Cylinder B. The internal pressure is 0.81 atm. otal Number of molecules = 11 Xe (red) + 17 Kr (Blue) = 28 molecules. Mole fraction of Kr = # of molecules of Kr / Total molecules = 17 / 28 = 17/28. Partial pressure of Kr = Mole fraction off Kr * Total Pressure = (17/28) * 0.81 atm = partial pressure of Kr = 0.49 atm
Which equation would best allow us to calculate molar mass from the gas density, pressure, and temperature?
Gas density:
Self-contained self-rescue breathing devices, like the one in the figure below, convert CO2 into O2 according to the following (balanced) reaction: How many grams of KO2 are needed to produce 835.0 L of O2 at 13.6 °C and 1.00 atm?
Given, Pressure of O2 = 1.00 atm Temperature of O2 = 13.6 oC = 286.75 oC volume of O2 = 835.0 L Now, Applying ideal gas equation for oxygen, we get, (1.00atm*835.0l)/(0.082atm*l/mol*k*286.75) =35.51mol o2 By stoichiometry of chemical relation, moles sKO2 needed=(4*(mole of reacted o2)/3 (4*35.51mol )/3 =47.35mol So, 47.35mol*71.1g/mol =3366.6g
An unknown solid has a molar heat capacity of Cp=15.5*J/mol×∘C How much heat (q) is absorbed when 1.00 mol of this unknown at -150.0 °C is heated to its melting point at -94.0 °C?
Given: Molar heat capacity of unknown solid: Cp= 15.5Jmol×C number of moles of unknown solid: n = 1.00mole Initial temperature: T1 = -150 oC Final temperature: T2 = -94 oC Heat absorbed = n × Cp × ( T2 - T1) Heat absorbed={1 mol×15.5Jmol×C×(−94−(−150))C} Heat absorbed = 868.0 J = 0.868 kJ
Calculate the heat capacity of a calorimeter if the combustion of 4.659 g of benzoic acid led to an increase in temperature of 7.148 °C.
Heat absorbed by calorimeter = Heat released by combustion = mass of benzoic acid x heat of combustion = 4.659g x 26.38 = 122.90 kJ Heat capacity of calorimeter = Heat absorbed by calorimeter/temperature change = 122.90kj/7.048c = 17.19 kJ/oC
Which of the following best defines heat?
Heat is the energy transferred between objects because of a difference in their temperatures.
The aromatic hydrocarbon cymene (C10H14) is found in nearly 100 spices and fragrances including coriander, anise, and thyme. The complete combustion of 1.608 g of cymene in a bomb calorimeter (Ccalorimeter = 3.640 kJ/°C) produced an increase in temperature of 19.35°C. Calculate the molar enthalpy of combustion of cymene (ΔHcomb) in kilojoules per mole of cymene.
Heat released by combustion = heat capacity of calorimeter x temperature change = 3.640 x 19.35 = 70.434 kJ Moles of cymene = mass/molar mass of cymene = 1.608/134.22 = 0.01198 mol Molar enthalpy of combustion ΔH = -heat released/moles of cymene = -70.434/0.01198 = -5879 kJ/mol Note that enthalpy is negative as reaction is exothermic and heat is released
The following reactions are carried out at atmospheric pressure (external) inside the chamber of a piston. Will the piston plunger have moved out (expanded) or moved in (contracted) after the reaction has occurred and the system was allowed to return to its initial temperature of 110°C?
N2(g)+3H2(g)2NH3(g) =The piston contracted.
Which of the following are used as units for pressure? Select all that apply.
Pa (Pascal) - A standard unit of pressure in the International System of Units (SI). Torr (Torr) - A unit of pressure often used in chemistry and related fields. atm (Atmosphere) - A unit of pressure commonly used in various applications, especially in the context of gases. kPa (Kilopascal) - A unit of pressure, equal to 1000 Pascals.
Calculate ΔE for a gas that releases 4.93 kJ of heat at the same time as it is compressed from 11.60 L to 5.40 L by a constant external pressure of 5.00 atm. Recall that 1 L•atm=101.32 J
Part 1 - Given ΔH=−4.93kJ initial volume -11.60L and final volume - 5.40L and 1L⋅ATM 101.32J with external pressure of 5.00atm. Work done - Pext−ΔV W=Pext ( Vf−Vi) q =- 4.93KJ = -4930J P = 5atm V1 = 11.6L V2 = 5.4L W = -p?v = -5atm*(5.4-11.6)L = 31L-atm = 31*101.32 = 3140.92J ?E = q + W = -4930+3140.92 = -1789J Explanation: As the value of change in enthalpy is given and work done is calculated by the formula after which the change in internal energy is also known with the formula which is ΔE=1.79KJ.
What properties of real gas molecules are associated with parameters a and b, respectively, of the van der Waals equation?
Property associated with a:=attractive forces between the gas molecules Property associated with b=volume of space occupied by the gas molecules
Which of the following is a state function associated with a thermodynamic system? Select all that apply.
So, internal energy(E), pressure(P) , volume(V) , temperature(T) are state functions.
Which of the following best defines work?
The energy required to move an object through a given distance
The temperature of a 100.0 g sample of water is raised from 30.0°C to 100.0°C. How much energy is required for the temperature change?
The equation to relate energy to a change in temperature is:$$q=ncpΔTq is the energy required, n is the number of moles of water, cp is the molar heat capacity for water, and ΔT is the change in temperature of the water.
What is the energy change when 43.0 grams of water cools from 94.0 °C to 24.5 °C?
The mass of water, m=43.0g The specific heat of water, CP=4.184Jg.K Therefore, the required energy change, Q=mCP(T2−T1) Q=43.0g(4.184Jg.K)(24.5-94.0)K Q=−12,503.884j Explanation: Here the negative sign indicates that the heat is given by the process. =-12500
A scuba diver releases a balloon containing 125.0 L of helium attached to a tray of artifacts at an underwater archaeological site (see photo). When the balloon reaches the surface, it has expanded to a volume of 442.0 L.
The pressure at the surface is 1.000 atm. What is the pressure at the underwater site? 442.0/125=3.536atm If pressure increases by 1.000 atm for every 10.00 m of depth, at what depth was the diver working? Assume the temperature remains constant. 3.536atm-1.000 2.536atm 10.00m/1.000atm*2.536atm =25.36m
Which of the following are used as units for volume? Select all that apply.
Units of volume include: milliliters (mL), liters (L), and cubic centimeters (cm3). You may also see cubic decimeters (dm3) in some instances. One cubic decimeter equals one liter.
A 100.0-mL flask contains 0.193 g of a volatile oxide of nitrogen. The pressure in the flask is 760 mmHg at 17°C. Is the gas NO, NO2, or N2O5?
Use the gas density equation to calculate ℳ d=mV=MPRT �=��=���� which rearranges to M=mRTVP �=�����Be sure to use units of volume in L, pressure in atm, and temperature in K. Solve M=(0.193 × 0.08206 L atm/(mol K) × 290 K)(0.100 L× 1.0 atm)=45.9 g/mol =(0.193 × 0.08206 L atm/(MOL K) × 290 K)(0.100 L× 1.0atm )=45.9 The molar masses of NO, NO2, and N2O5 are 30.01, 46.01, and 108.02 g/mol, respectively. The gas in the flask is NO2.
Calculate the amount of energy needed to convert 13.9 g of water at 30.0 °C to steam at 100.0°C.
Water will undergo the following transition : Water (28C)⟶Water (100C)⟶Steam (100C) Water to Water : Heat=Mass×Specific heat×ΔT=13.9×4.184×(100−30.0)=J=4071j *1kJ/1000j =4.07kj Water to Steam : Heat=Mass×Heat of Vaporization=13.9×2,260=31414 J*1 kJ/1000j =31.41kj therefore, Total Heat=4.07kj+31.41kj=35.5 kJ
Determining energy of a system (memorize: ΔE = q + w)
______ heat, work, endothermic, exothermic... and signs ______ units of energy (J, kJ, Calories and interconversion)
Chemical energy
______ stoichiometry & sign of ΔH in a chemical reaction ______ Constant pressure calorimetry (memorize: qp = CS x m x ΔT) ______ Constant volume calorimetry (memorize: qv = Ccal x ΔT)
Is the following process exothermic or endothermic? ice cubes melt
endothermic
For the heating curve shown below, select all the processes that occur as you move leftward from the gas state at Point A toward Point B.
freezing,condesation, liquid is cooled ,freezing Processes occurring at a constant temperature are phase changes. These are the horizontal line segments on the heating curve. The angled line segments represent temperature change, either heating or cooling the substance depending on the direction you are moving on the heating curve.Point A is in the gas portion of the heating curve, and Point B is in the solid phase at the freezing point. Moving between these two points requires four different processes: - cooling the gas (decrease in T);- changing from gas to liquid (condensation);- cooling the liquid (decrease in T);- changing from liquid to solid (freezing).
A bomb calorimeter has a heat capacity of 3.53 kJ/°C. A sample of cinnamaldehyde is ignited in the calorimeter, resulting in a temperature increase from 24.53°C to 31.23°C.
heat cap=3.53 kJ/°C. temp change=31.23c-24.53c=6.7c heat created= heat cap*DT =3.53 kJ/°C*6.7 =23.7kj
PATH FUNCTIONS
heat flow(q) , work done (w)
A mixture of 0.158 moles of C is reacted with 0.117 moles of O2 in a sealed, 10.0 L vessel at 500.0 K, producing a mixture of CO and CO2. 3C(s)+2O2(g)2CO(g)+CO2(g) The total pressure is 0.696 atm. What is the partial pressure of CO?
mass 0.158 molesC mass 0.117 moles of O2 0.105 0.012 0.0525 =0.1695
Generating hydrogen from water or methane is energy-intensive. A non-natural enzymatic process has been developed that produces 12 moles of hydrogen per mole of glucose by the reaction: What volume of hydrogen could be produced from 256 g of glucose at STP? L
mass=given mass/molar mass 256 g/180.16gmol^-1 1.420mol According to the balanced equation: 1-mole of the glucose produces =12-molof teh hydrogen gas 1-mole of hydrogen gas=12-mol of hydrogen gas/1-mol of glucose×(1.420 mol of glucose) 1-mol of hydrogen gas =17.05mol of h2 the volume taken at stp=moles*22.4l volume=17.05*22.4l =381.95l
The graph below has three lines indicating average, most probable, and root-mean-square speeds. Using the formulas for each speed, determine which line and speed correlate.
most probable>average speed>root-means square speed
A gas mixture contains 0.700 mol of N2, 0.150 mol of H2, and 0.400 mol of CH4. Calculate the pressure of the gas mixture and the partial pressure of each constituent gas if the mixture is in a 12.0 L vessel at 27.00°C. total pressure in the vessel pressure of H2 pressure of N2 pressure of CH4
n2=0.700mol h2=0.150mol ch4=0.400mol 12.0l 27.00c=300.15k total mol=1.25moll =2.57 ph2=(0.150/1.25)*2.57=0.308 pn2= (0.700/1.25)*2.57=1.439 pch4=(0.400/1.25)*2.57 =0.822
Use the data table below to answer the following questions. What is the standard heat of formation for N2O(g)? What is the standard heat of formation for H2O(g)?
n2o(g)=62.1kj/mol h2o=-241.8kj/mol
Assume that the kernels of unpopped popcorn shown in the image constitute a thermodynamic system. Sort the quantities w, q, and ΔE for this system during the popping process according to their signs. (Hint: Assume that there is a large volume change when popcorn pops.)
negative =ΔE,w positive= q
The value of ΔHcomb for dimethylphthalate (C10H10O4) is 4685 kJ/mol. How much heat was released during the combustion of 0.950 g of dimethylphthalate?
number of moles=0.800g/194g/mol =0.004897mol heat released ΔHcomb x number of moles 4685kj/mol x 0.004897mol =22.9kj
Helium effuses 3.16 times as fast as which other noble gas?
r1/r2 = sq. root of M2/M1 3.16/1.00 = sq.root of M2/4.00 (3.16/1.00)^2 = M2/4.00 9.98 = M2/4.00 M2 = 39.9 that is the gas is Argon
How would you combine reactions A-C, shown below, to obtain the overall reaction: NH3(g)+BH3(g)+O2(g)2H2O(g)+HBNH(s) Please select all that apply.A. 2H2(g)+O2(g)2H2O(g) B. H3BNH3(s)NH3(g)+BH3(g) C. H3BNH3(s)2H2(g)+HBNH(s)
reverse b and add resultng rquations together Ammonia, BH3, and oxygen gas are all reactants and need to be on the left side of the reaction arrow. H3BNH3(s) and hydrogen gas do not appear in the overall reaction, so we need to be sure they are on the reactants side in one reaction and the products side in another so they will cancel out.SolveIn order to get ammonia and BH3 on the reactants side and have H3BNH3 cancel out, we can reverse equation B. Once we have reversed B, if we add all three equations together we will get the desired overall reaction. Think About ItΔH reaction is calculated by the products - reactants. If we reverse the equation, we would be swapping the reactants and products. This would result in a change in the sign of ΔH.
How much PΔV work does a gas system do on its surroundings at a constant pressure of 1.15 atm if the volume of gas triples from 305.0 mL to 9.150×102 mL? Express your answer as an absolute value in L·atm
vi 305.0ml*1L/1000ml=0.305L vf 9.150×10^2 mL**1L/1000ml=0.915L Now we can calculate the change in volume (ΔV): ΔV=V2−V1=0.915L−0.305L=0.61L Plugging in the values into the formula, we get: Work=PΔV=(1.15atm)(0.61L)=0.7015l·atm
How much PΔV work does a gas system do on its surroundings at a constant pressure of 1.15 atm if the volume of gas triples from 305.0 mL to 9.150×102 mL? Express your answer as an absolute value in joules (J).
vi 305.0ml*1L/1000ml=0.305L vf 9.150×10^2 mL**1L/1000ml=0.915L Now we can calculate the change in volume (ΔV): ΔV=V2−V1=0.915L−0.305L=0.61L Plugging in the values into the formula, we get: Work=PΔV=(1.15atm)(0.61L)=0.7015l·atm To convert L·atm to joules (J), we use the conversion factor: 1L·atm=101.325J So, the work done by the gas system on its surroundings is: |Work|=0.7015l·atm×(101.325JL·atm)=71.08j
Calculate ΔE for a system that absorbs 579.0 kJ of heat from its surroundings and does 547.0 kJ of work on its surroundings.
ΔE=q+w q= heat -If the system losing heat q become negative, +If the system gaining heat q become positive. w= work If the system doing work (work done by the system) w become negative, +If the surrounding doing work on the system w become positive. q = absorbs heat, so it is positive = 579.0 KJ w = work is done by system on surrounding, so it is negative = -547.0 KJ ΔE=q+(−w)=579.0+(−547.0)=32kJ
Which of the following equations accurately expresses the first law of thermodynamics?
ΔEsys+ΔEsurr=0
What happens to the magnitude and sign of the enthalpy change when a process is reversed?
ΔH changes sign, but keeps its absolute value.
Adding nitric acid to water is quite exothermic.If the ΔH°soln of HNO3 is -33.3 kJ/mol, then how much heat is evolved by dissolving 0.150 mol HNO3 in 100.0 mL of water?
ΔHsol for HNO3 is −33.3 kJmol−1 need to convert this value to joules per mole: ΔHsol=−33.3kJmol×1,000JKJ=−33,300 Jmol−1 Now you can plug in the values: n is 0.150mol ΔHsol is −33,300 Jmol−1 n is 0.180molΔHsol is −33,300 Jmol−1 q=n×ΔHsol =0.150 mol×(−33,300 Jmol−1)=−4,995 J .Since the heat is evolved, the negative sign indicates that heat is released to the surroundings. Therefore, the heat evolved by dissolving 0.180mol of HNO3 in 100.0mL of water is−4995J.
Calculate ΔE for the following situations: q = -625.0 J; w = -425.0 J
∆E = q+W -625.0 J+ -425.0 J=-1050.0j
Calculate ΔE for the following situations: q = 115.0 J; w = -30.00 J
∆E = q+W equation 1 115.0+-30.00=85
