CBT4 Flashcards
Passage II. 8. When reaction IV is in a state of equilibrium, which of the following changes will cause more NH3 to form???
a) addition of H2 b)a decrease in pressure c)an increase in temp d)addition of catalyst
a) Addition of H2.
When reaction 4 is in equilibrium, LE CHATELIER principles are applicable. If the concentration of either the reactants, H2 OR N2, is increased, the position of the equilibrium shifts to the right, and more NH3 forms. A DECREASE IN TOTAL PRESSURE OF REACTANTS AND PRODUCTS SHIFTS THE EQUILIBRIUM TO THE SIDE WITH THE MORE GASEOUS MOLECULES.
Passage IV. 23. At the stoichiometric (equivalence) point in a titration of benzoic acid with NaOH (aq) shown by Eq. 3, the pH is:
a)<a>4 and 7</a>
D)>7
THE PH AT THE EQUIVALENCE POINT IN ANY TITRATION IS THE pH of the SALT SOLUTION FORMED. At the equivalence point, the solution contains BENZOATE and water. Passage states benzoic acid is a WEAK ACID; therefore its conjugate base, the benzoate anion, is a stronger base that OH- and reacts with water to produce OH- and undissociated benzoic acid. RESULTS IN SOLUTION WITH pH>7
ST.Alone 17. The table above gives pressure measured at various depths below the surface of a liquid in a container. A second liquid, whose density is TWICE that of the first liquid, is poured into a second container. Similar pressure measurements are taken for the second liquid at various depths below the surface of the second liquid. What is the pressure at the depth of 10 cm for the second liquid?
a) 250 N/m^2 b)450 N/m^2 c)850 N/m^2 d) 1650 N/m^2
C) 850 N/m^2
The pressure in a liquid due to the gravitational force of the liquid above a given depth is proportional to the density and the depth. Although the pressures shown in the table change linearly with depth (and increase of 200 N/m^2 per 5 m of depth), the data also imply an extra pressure of 50 N/m^2 at zero depth. For the same depth, the liquid with twice the density will create twice as much pressure as the first liquid, but the zero depth pressure must be added to get the total pressure.
(2x400+50) =850
)
St. Alone 32. Suppose that CH4 (g) reacts completely with O2 (gas) and H2O (g) with a total pressure of 1.2 torr. What is the partial pressure of H2O (g)?
a) 0.4 torr b) 0.6 torr c) 0.8 torr d)1.2 torr
C) 0.8 torr
The balanced equation for the complete combustion of CH4 is
CH4 + 2O2 —> CO2 + 2H2O
The pressure of the gaseous products is 1.2 torr. For every three product molecules, 2 are water, so the partial pressure of water is 2/3 the total pressure, because the total pressure is function of the total number but not kind of molecules. Two thirds of 1.2 is 0.8.
Ptotal=PCO2 + PH2O.
Passage V. 27. Which of the following expressions gives the MOLE FRACTION of KNO3 in the solution prepared by the student?
a) (906.1/18.0) / ((226.5/101.1)+(906.1/18)) b)(226.5/101.1) / (226.5/101.1)+(906.1/18) c) (906/18) / (226.5/101.1) d) (226.5/101.1)/(906.1/18.0)
B)
Mole fraction of KNO3 in the solution is the number of moles of KNO3 divided by the total number of MOLES of all chemical species in the solution.
PV. 28. Crystals precipitate when each of the following compounds is added to a saturated solution of KNO3 (aq) EXCEPT:
a) NH4NO3 (s) b)Ca(NO3)2 (s) c) NH4Cl (s) d)KCl (s)
c) NH4Cl (s)
A saturated solution of KNO3 contains the MAXIMUM NUMBER OF K+ and NO3- ions the solution can hold w/o precipitating solid KNO3. The addition of any salt that ADDS EITHER OF THESE TWO IONS WILL CAUSE KNO3 to form (Crystals to precipitate). This is the COMMON-ION EFFECT. The only answer choice that does not contribute common K+ or NO3- is NH4Cl.
PV.30. How many grams of KNO3 are in 100 mL of the student prepared solution of KNO3 (aq)?
a) 11.33 g b)22.41 c)22.65 d)34.0
c) 22.41 g
From table 1, there is 226.5 g KNO3, the solute, per liter of soltion. In 100 mL, there is one tenth the mass, so 22.65 g.
226.5 g/L x 0.1 L = 22.65 g
PV. 31 What is the approximate number of potassium ions in the student prepared solution of KNO3?
A) 10^9 B) 10^16 C) 10^20 D) 10^24
d) 10^24
There is one mole of K+ IN EACH MOLE OF KNO3. From table 1, the molarity of the student solution is 2.241 M. The solution contains 2.241 moles of K+ or 2.241 x Avogadro’s number of K+ions.
2.241 x 6.02e23= 1.2e24 = 10^24
PVI. 39 What kind of object is emitted in the decay of 40 K?
a) gamma ray b) alpha particle c) electron d) positron
D)Positron
The 40 K to 40 Ar radioactive decay does not change the ATOMIC MASS (40) of the nucleus, but DOES CHANGE its ATOMIC NUMBER (19 to 18). These conditions point to a POSITRON as the emitted object.
POSITRON EMISSION: A PROTON IS CHANGED INTO A NEUTRON, with expulsion of POSITRON.
PVII.43. When the number of photons incident on the cathode with energies above the value of the work function increases, which of the following quantities also increases?
a) number of electrons ejected b)potential energy of each ejected photon c) magnitude of electric field between the electrodes d) speed of electrons at the anode.
a) number of electrons ejected.
Then number of incident photons affects only the number of electrons, not their energies. The electron energies depend on PHOTON ENERGY, the cathode work function, and the potential difference between the cathode and anode.
ST Alone 51. Suppose a certain far-sighted person can see objects clearly no closer than 300 cm away. What is the minimum distance from a plane mirror such a person must be to see his reflection clearly?
a) 75 cm b) 150 cm c) 300 cm d) 600 cm
B) 150 cm
A PLANE MIRROR PRODUCES AN IMAGE BEHIND ITS PLANE AT A DISTANCE EQUAL TO THE OBJECT DISTANCE IN FRONT OF THE PLANE. If the object plus image distances must be at least 300 cm, then the plane of the mirror must be half this distance, 150 cm.
PIII.15. An astronomer observes a hydrogen line in the spectrum of a star. The λ of hydrogen in the lab is 6.563e-7 m but the λ in the star’s light is measured at 6.56186e-7 m. Which of the following explains this discrepancy?
a) star is moving away from the earth b) wavelength of light that the star is emitting changes constantly c) frequency of light that the star is emitting changes constantly d) the star is approaching Earth
d) The star is approaching Earth
The WAVELENGTH OF THE LIGHT detected from the star is SMALLER than the lab value on Earth. This implies a DOPPLER SHIFT (blue shift) associated with approaching relative velocity between star and Earth.
PI.4. After a block began to slide, how did its speed cary with time? (Note: Assume that the tension and kinetic friction forces on the block were constant in magnitude)
a) it was constant in time b) it increased exponentially with time c) it was first constant, then increased linearly with time d) in increased linearly with time
D) It increased linearly with time
The coefficient of kinetic friction is ALWAYS lower than that of static friction. Therefore there is a NET ACCELERATING FORCE on the block once it starts to slide. A constant force on a mass produces a constant acceleration, thus the velocity of the block increases linearly with time.
ST. ALONE 34. The fundamental, resonant λ of a pipe open at BOTH ENDS that is 1m long and 0.1 m in diameter is:
a) 0.1 m b) 0.2 m c) 1.0 m d)2.0 m
d) 2.0 m
Pipes and tubes have their resonant wavelengths when standing waves develop. An open pipe has its fundamental resonant wavelength at TWICE the length of the pipe. Both ends have displacement ANTINODES (maximum amplitudes) with a node in the middle of the pipe. Thus, the pipe is half a wavelength long. The resonant wavelength is INDEPENDENT OF DIAMETER.
PAGE 246 λ= 2L/n
PV124. The two primary factors that normally determine the level of blood pressure are: a)blood concentration of L-NMMA and NE b) cardiac output and the resistance to blood flow c) the blood volume and the amount of L-arginine in the diet d)heart rate and the cardiac stroke volume.
B) CO and the resistance to blood flow
Two factors that normally determine the blood pressure are the cardiac output and the resistance to blood flow. Cardiac output (SV X HR) determines the amount of blood pumped into the system by the heart per unit time. The resistance to blood flow is primarily determined by the caliber of the small arteries, arterioles, and precapillary sphincters. Thus BLOOD PRESSURE = TOTAL PERIPHERAL RESISTANCE X CO, a relationship analogous to OHM LAW.
