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DAT ACTUAL EXAM TEST PAPER FULL QUESTIONS CORRECT RESPONSES.pdf

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DAT ACTUAL EXAM TEST PAPER FULL QUESTIONS CORRECT RESPONSES




Question:
If the temperature drops by 18o Fahrenheit in a given day, how much did it drop in degrees Celsius?
a) -14 b) 10 c) 32.4 d) -7.78 e) 25.2.

Answer:
b) 10 The ratio of "change in degrees Fahrenheit" to "change in degrees Celsius" is simply 9 to 5,
and therefore, (18×5) / 9 = 10. If you were converting a temperature of 18o Fahrenheit to degrees
Celsius, then it would be ((18-32) × 5) / 9 = -7.78o Celsius.



Question:
Which of the following will not dissolve in water? A. Salt, NaCl B. Ethanol, CH3CH2OH C.
Benzene, C6H6 D. Hydrogen chloride, HCl E. Acetic acid, CH3COOH.

Answer:
C. Benzene, C6H6. The law of solubility is that "Like dissolves Like", that is non- polar solvents
dissolve non-polar substances and polar solvents dissolve polar and most ionic substances. Water is
a polar solvent thus: Salt, NaCl - ionic and very soluble in water, sea water. Ethanol, CH3CH2OH -
polar molecular (-OH) will dissolve in water, beer/wine/spirits. Benzene, C6H6 - non-polar
molecular will not dissolve in polar water. Hydrogen chloride, HCl - polar molecular will dissolve
in water, hydrochloric acid. Acetic acid, CH3COOH - polar molecular (-COOH) will dissolve in
water, vinegar.



Question:
Which of the following structures of the skeletal system is innervated? A. Hair B. Nails C. Cartilage
D. Osteoclasts E. Haversian canals.

Answer:
E. Haversian canals Innervated refers to the structure having nerves and blood vessels that run
through it. Nails and hair are consider dead/dying cells and would not have blood vessels that run
through the exterior nail structure. Cartilage is not innervated which attributes to the long healing

,time of sprained joints. Osteoclasts are cells that help to break down bone cells and these cells
themselves do not have blood vessels or nerves. The Haversian canals are a network of opening in
the bone that allow for nerves and blood vessels to weave in and out of the bone.



Question:
Two trains are approaching each other from opposite directions with the speeds 10 miles per hour
and 20 miles per hour, and heading for a collision. If they are a quarter mile apart, how long will it
take for them to collide? A. 45 seconds B. 20 seconds C. 30 seconds D. 40 seconds E. 1 minute.

Answer:
C. 30 seconds The speed at which the two trains are closing the gap between them is 10 + 20 = 30
mph. At the is speed, it will take (0.25/30) × 3600 = 30 seconds to close a gap of 0.25 miles.



Question:
Which of the following statements is correct? A. High molecular dipole moments are characteristic
of nonpolar molecules B. Polar solutes tend to be more soluble in nonpolar solvents C. The polarity
of a molecule is dependent on its 3-D structure D. All polar molecules are capable of hydrogen
bonding E. London dispersion interactions are usually stronger than dipole-dipole interactions.

Answer:
C. The polarity of a molecule is dependent on its 3-D structure A molecule's dipole is an electric
dipole with an inherent electric field (not be confused with a magnetic dipole which generates a
magnetic field). Molecules can have dipole moments due to non- uniform distributions of positive
and negative charges on the various atoms. For diatomic molecules there is only one (single or
multiple) bond so the bond dipole moment is the molecular dipole moment, with typical values in
the range of 0 to 11 D. For polyatomic molecules there is more than one bond, and the total
molecular dipole moment may be approximated as the vector sum of all individual bond dipole
moments.



Question:
If an unknown compound is found to have an empirical formula of CH4, which of the following is
the molecular formula if the molar mass was determined to be 128 g/mol? A. CH4 B. CH12 C.
C8H12 D. C8H32 E. C12H16.

Answer:

,D. C8H32 To determine the molecular formula using an empirical formula, the ratio of the
molecular mass to the empirical mass must be calculated. Dividing the molecular mass of 128 grams
by the empirical mass of ~16 grams (C = 12g + H = 1g, but there are 4 hydrogens in the compound)
gives you approximately 8. This indicates that the empirical subscripts must be multiplied by 8 to
result in the molecular formula of the unknown compound, C8H32.



Question:
Which of the following molecules is amphipathic? A. water B. glucose C. steroids D. triglycerides
E. phospholipids.

Answer:
E. phospholipids An amphipathic molecule has both a hydrophilic ("water loving") and a
hydrophobic ("water fearing") region. Water is hydrophilic, whereas a triglyceride (or lipid) is
hydrophobic. Glucose is also hydrophilic. Steroids are derived from lipids and also would be
hydrophobic. A phospholipid is made of a phosphate group (hydrophilic) and two fatty acid chains
(hydrophobic) and is amphipathic.



Question:
From a group of 5 men and 7 women, in how many different ways can a team of 2 men and 3
women be formed? A. 95040 B. 792 C. 350 D. 4200 E. 18.

Answer:
C. 350 2 men can be picked from a group of 5 in (5|2) = 10 different ways, and 3 women can be
picked from a group of 7 in (7|3) = 35 different ways. Therefore, the total number of ways a team
can be formed is 10 × 35 = 350. Extended explanation: If you got D. 4200, you interpreted this
problem as a permutation, not as a combination. We do not care what order the men are picked in or
what order the women are picked in, all we care is that there is a team of 2 men and a team of 3
women. Permutation equation: P(n,r) = n! / (n-r)! Combination equation: C(n,r) = n! / r!(n-r)! Where
n = total number of the group and r = the number of people we want. Hence, n = 5 men r = 2 men
We have 10 total possible combinations using the above formula. The same goes for women. n = 7
women r = 3 women We have 35 total possible combinations using the above formula. Multiplied
together, we have 350 possible combinations.



Question:
How many degrees of unsaturation are present in the anxiolytic drug Valium (diazepam), molecular
formula C16H13ClN2O? A. 8 B. 9 C. 10 D. 11 E. 12.

, Answer:
D. 11 Degree of Unsaturation (or index of hydrogen deficiency, IHD), or rings plus pi bonds, lets
you determine the number of rings, double bonds and triple bonds present in the compound if the
molecular formul is known. It does not give the exact number of rings or double or triple bonds, but
rather the sum of the number of rings and double bonds plus twice the number of triple bonds.
Degree of Unsaturation = C - (H/2) - (X/2) + (N/2) +1 (C=#carbon, H=#hydrogen, X=#halogen,
N=#nitrogen) = 16 - (13/2) - (1/2) + (2/2) + 1 = 11 Extended Explanation First use the saturated
formula CnH(2n+2) which will get you to C16H34 Now lets deal with the Cl. How does it play a
role in this molecule? It connects with a carbon and TAKES the place of a H. So the saturated
molecule should have 1 less Hydrogen from what we predicted. Now we have C16H33 Now the two
nitrogens. Lets assume they are amines for the saturated solution. They connect with the carbon and
TAKE the place of a hydrogen BUT add on TWO hydrogens (NH2). Net gain of 1 hydrogen per N.
So the saturated molecules would have 2 more H of the current formula we predicted. Now we have
C16H35. All you have to do now is subtract the H difference divided by two. 35-13 = 22/2 = 11
degrees of unsaturation



Question:
What is the final oxidation number of carbon in the following chemical reaction? 6CO2 + 6H2O ->
C6H12O6 + 6O2 A. -4 B. -2 C. 0 D. 2 E. 4.

Answer:
C. 0 In the equation, carbon is used from carbon dioxide to form glucose, C6H12O6. The final form
of carbon is in glucose. In this compound, the oxygens each have an oxidation number of - 2 and the
hydrogen also have a fixed oxidation number of +1 each. Therefore, the sum of hydrogen's
oxidation numbers is +12 and the sum of oxygen's oxidation numbers is -12. Since glucose is a
neutral compound the sum of all the oxidation numbers of all the elements in the compound must be
equal to zero. So, carbon must have an oxidation number of 0 because the sum of the hydrogen and
oxygen oxidation numbers is already equal to zero.



Question:
A particle that can bind in the active site of an enzyme, preventing the binding of the substrate is
referred to as which of the following? A. positive inhibitor B. negative inhibitor C. competitive
inhibitor D. non-competitive inhibitor E. regulator proteins.

Answer:
C. competitive inhibitor If the particle can bind to the active site, it is competing for the active site
with the substrate. A non-competitive inhibitor would bind to an allosteric site on the enzyme,

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