5) A neutralization reaction between KOH (ag) and H2S04 (aq) would give which tw

Question

5) A neutralization reaction between KOH (ag) and H2S04 (aq) would give which two products? A) H20 () and H2S (g) B) H20 (1) and KSO4 (aq) C) H20 () and K2504 (aq) D) SO2 (g) and KH2 (g) E) none of the above Answer: 6) A 25.0 ml sample of 0.105 M HCI was titrated with 315 mL of NaOH. What is the concentration of the NaOH? A) 0.0833 MM B) 0.132 M C) 0.105 M D) 0.075 M E) none of the above Answer: 2 Hox many sramsof Lif would be present in 575 mlL of 0z30 M LIF solution? A) 11.2 B) 0.0338 C) 1.12 x 104 D) 19.9 E) 33.8 Answer: 8) How many grams of barium sulfate are produced if 2534 mL of 0113 M BaC12 completely react given the reaction Baa2(aq) + Na2SOg(aq) BaSQ4(s) + 2NaCl (aq) A) 5.90 B) 26.3 C) 1039 D) 0.668 E) none of the above Answer: 9) What is the pressure of a 3.00 L gas vessel that has 18.0 grams of helium at 5C? (R-0.0821 L atm/ mol K) A) 147 atm B) 36.7 atm C) 32.6 atm D) 1.81 atm E) none of the above Answer

Explanation / Answer

5.

H2SO4(aq) + KOH(aq) ------>K2SO4(aq) + 2H2O(l)

Ans = (C)

6.

as both are mono basic and mono acidic

M1V1 = M2V2

M1 and M2 = molarity

V1 and V2 = volume

25 mL x 0.105 M = M2 x 31.5 mL

M2 = 0.0833 M

ans = (A)

7.

moles = molarity x volume in litres = 0.750 M x 0.575L = 0.43125 mole

molar mass of LiF = 25.939 g/mol

mass = moles x molar mass = 0.43125 mol x 25.939 g/mol = 11.186g

~ 11.2g

ans = (A)

8.)

BaCl2 + Na2SO4 ------------> BaSO4 + 2 NaCl

1 mol BaCl2 forms 1 mol BaSO4

moles of BaCl2 = molarity x volume in litres = 0.113M x 0.02534L = 0.00286342 mol

moles of BaSO4 formed = moles of BaCl2 reacted = 0.00286342 moles formed

molar mass of BaSO4 = 233.38 g/mol

mass obtained = moles x molar mass = 0.00286342 mol x 233.38 g/mol = 0.6683 g

~0.668g

ans = (D)

9.

moles of Helium = mass / molar mass = 18g / 4g/mol = 4.5 moles

25 oC = 25 + 273 = 298 K

according to ideal gas P = nRT/V

P = 4.5 mol x 0.08206 atm.L/mol.K x 298K / 3.00L = 36.68 atm ~ 36.7 atm

ans = (B)

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