Prepare for the Dental Admission Test (DAT) with this practice test and answers. This guide covers biology, general chemistry, organic chemistry, and perceptual ability.
Q: If 1 mole of N2 and 1 mole of H2 aremixed and allowed to react according tothe equation N2 + 3H2 2NH3. Whatis the maximum number of moles of NH3that could be produced?
Answer: This is a limiting reagent problem. There is 1 mole of nitrogen and 1 mole ofhydrogen, however according to the chemical equation, 1 mole of nitrogen is neededfor every 3 moles of hydrogen. Thus, the limiting reagent is hydrogen. The amount ofammonia that can be produced is determined as follows:1 mole H2 x (2 ššš ššššššš/3 ššš āš¦šššššš) = 2/3 moles of ammonia
Q: A flask weighs 95 g when empty. Whenfilled with 200 mL of a certain liquid, theweight is 328 g. What volume (inmilliliters) would 1,000 g of the liquidoccupy?
Answer: B – The mass of the liquid can be determined by subtracting 328g – 95g. Aproportion can be written relating the mass of the liquid to the volume the liquidoccupies.200 ššæ/328šā95š= š„ ššæ/1000š, then cross multiply to solve for the unknown volume, resulting in:(200)(1000) = (328-95) (xmL), dividing to solve for x, gives: (200)(1000)/(328ā95)
Q: How many grams of NaOH (40 gmol-1)are there in 250 mL of 0.4 M NaOHsolution?
Answer: Molarity = ššššš šš š ššš¢š”š / ššš”ššš šš š ššš¢š”ššš0.4 M = ššššš šš šššš» 0.250šæ(0.4)(0.250) = moles of NaOH(0.4)(0.250) = 0.1 moles NaOH x 40š šššš»1 šššš šššš» = 4 g NaOH
Q: In which reaction is H2O considered to be acting as an acid?
Answer: NH3 + H2O <->NH4+ + OH-
Q: The concentrations of silver ion andchloride ion in an aqueous solution in equilibrium with solid silver chloride are 1.0 x 10-6 M. What is the value of Ksp for AgCl equal to?
Answer: For the decomposition of AgCl, the following equation would result:AgCl(s) Ag+1(aq) + Cl-1(aq) The subsequent solubility product expression (Ksp) would be: Ksp = [Ag+1][Cl-1]. If theconcentration of each of the ions is 1.0 x 10-6 then the Ksp would be:[1.0 x 10-6][1.0 x 10-6] or [1.0 x 10-6]^2.
Q: Reducing agent
Answer: gets oxidized so should show an inccrease in the charge (+) as it means it lost an e-. Reducing agent is in the reactants not the products.Cr2O + 14H+ + 6I-> 2Cr3+ +7H2O +3IThe The iodide ion, I-, has an oxidation number of -1, but is converted to I2 which has an oxidation number of 0. The iodide ion is thereducing agent (loses electrons and becomes more positive) and is oxidized in thechemical reaction.
Q: How do you derive the EĀ° of an equation
Answer: you must add up both equations to make sure they give the overall equtation asked for:Cl2(g) + 2e-> 2Cl-(aq) EĀ°= +1.36vCu2+(aq) + 2e-> Cu(s) EĀ°= +0.34v======================Cu2+(aq) + 2Cl->(aq) Cu(s) + Cl2(g)Therefore flip the first one and add to -1.36 + .34 = -1.2
Q: beta decay, alpha decay, positron emission, e- capture, gamma decay
Answer: alpha decay- 2 protons and 2 neutrons lost..Beta decay (e- emission)- p+ transformed into Neutron 1 e- lostPositron emission- 1 P+ is converted into neutron and 1e+ is released, has a +1/2 spin.
Q: what has to be done in a nuclear fission equation?
Answer: Both products and reactants superscripts/subscripts (top/bottom) have to equal each other on both sides.