Chem 30A Test 3 Ch 5,6,7,13,14,15 V1 Name 4· Some e commercial antacid tablets c

Question

Chem 30A Test 3 Ch 5,6,7,13,14,15 V1 Name 4· Some e commercial antacid tablets contain Caco, which neutralizes excess HCI in t and 1.02 atm pressure when 80mL of 0.050 MHCI reacts completely with Caco, in an antaci n dioxide and water. How many ml of co, gas is produced at 37°c antacid tablet? (4) (SET THE PROBLEM AND SHOW ALL WORKING) Caco, (s) + 2HCI (aq) CaCl2 (aa) + co, (g) + H,0) 5. Neon gas inside a gas cylinder with a volume of 1.0L is exerting a pressure of 101 kP 22"C. What will be the final volume of Neon, if all of this gas another container where it exerts a pressure of 1.3 kPa at 19°C. (Do not change any other units for pressure). (4) (WRITE THE FORMULA) is allowed to expand in PART B: Write the answer and mark the scantron. (10 X 3 30) 1) Which of the following molecules can form hydrogen bonds? A) CH4 B) NaH C) NH3 D) BH3 E) HI Answer 2) CsfHi does not dissolve in water because A) CsHs is polar B) CsaHu is nonpolar C) water is nonpolar. D) water is saturated E) CeHx is hydrated Answer Response on this test is a TRUE Reflection of my knowledge and understanding of the 2 concents in the chapters sienl

Explanation / Answer

4) The balanced stoichiometric equation for the reaction is given. As per the stoichiometric equation,

2 mole HCl = 1 mole CO2

Mole(s) of HCl used in the reaction = (80 mL)*(1 L/1000 mL)*(0.050 M)*(1 mol L-1/1 M) = 0.0040 mole.

Mole(s) of CO2 produced as per the balanced equation = (0.0040 mole HCl)*(1 mole CO2/2 moles HCl) = 0.0020 mole.

Use the ideal gas law: P*V = n*R*T where P = 1.02 atm, n = 0.0020 mole, R = 0.082 L-atm/mol.K and T = 37°C = (37 + 273) K = 310 K.

Plug in values and obtain

(1.02 atm)*V = (0.0020 mole)*(0.082 L-atm/mol.K)*(310 K)

====> V = (0.0020 mole)*(0.082 L-atm/mol.K)*(310 K)/(1.02 atm) = 0.04984 L = (0.04984 L)*(1000 mL/1 L) = 49.84 mL

The volume occupied by the evolved CO2 gas is 49.84 mL (ans).

5) Use the combined gas law

P1*V1/T1 = P2*V2/T2

where P1 = 101 kPa

V1 = 1.0 L

T1 = 22°C = (22 + 273) K = 295 K

P2 = 1.3 kPa

and T2 = 19°C = (19 + 273) K = 292 K

Plug in values and get

(101 kPa)*(1.0 L)/(295 K) = (1.3 kPa)*V2/(292 K)

====> V2 = (101 kPa)*(1.0 L)*(292 K)/(1.3 kPa).(295 K) = 76.9022 L 76.90 L

The volume occupied by the neon gas at 1.3 kPa pressure and 19°C is 76.90 L (ans).

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