The delta H for the solution process when solid sodium hydroxide dissolves in wa

Question

The delta H for the solution process when solid sodium hydroxide dissolves in water is 44.4 kL/mol. When a 13.5 g sample of NaOH dissolves in 250.0 g of water in a coffee-cup calorimeter, the temperature increases from 23.0 degree C to _______ degree C. Assume that the solution has the same specific heat as liquid water i.e 4.18 J/g-k 24.0 degree C 35.2 degree C 40.2 degree C 37.0 degree C Given the data in the table below, delta H^+ for the reaction 4NH_2(g) + 5O_2(g) right arrow 4NO(g) + 6H_2O(l) is ________ kJ -1892 -1540 -150 -1172 For which one of the following reactions is delta H^0_rxn equal to the heat of formula of the product 12C(g) + 11H_2(g) + 11)(g) right arrow C_7H_25O_2(g) P(g) + 4H(g) + Br(g) right arrow PH_4Br(l) N_2 (g) + 3H_2(g) right arrow 2NH_3(g) (1/2)N_2(g) + O_2(g) right arrow NO_2(g) The molecular weight of a gas that has a density of 5.75 p/L at 273 K and 1.00 ATM is ______ g/mol 129 141 3.90 5.76 One significant difference between gases and liquids is that _______ gases are always mixtures a gas is made up of molecules a gas assumes the volume of its container a gas may consists of both elements and compounds The thermal decomposition of potassium chloride can be used to produce oxygen in the laboratory 2KClO_3 (s) right arrow 2KCl(s) + 3O_2(g) What volume (L) of O_2 gas at 396K and 1.00 atm pressure is produced by decomposition of 7.5g of KClO_2(s) (123 g/mol) 3.7 11 7.5 2.2 _________ delta compounds to an ________ process negative, endothermic negative, exothermic positive, endothermic positive, exothermic

Explanation / Answer

32)   moles of NaOH = 13.9 / 40 = 0.3475

heat evolved = 0.3475 x 44.4

heat evolved = 15.429 kJ

now

total mass = 250 + 13.9 = 263.9 g

now

heat = m x s xdT

15.429 x 1000 = 263.9 x 4.18 x ( T -23)

T = 37 C

so

the answer is D) 37 C

33)

4NH3 + 502 --> 4 NO + 6H20

dHrxn = dHfo products - dHfo reactants

so

dHrxn = 4 x dHfo N0 + 6 x dHfo H20 - 4 x dHfo NH3 - 5 x dHfo 02

dHrxn = ( 4 x 90) + ( 6 x -286) - ( 4 x -46) - 0

dHrxn = -1172 kJ

so

the answer is D) -1172

34) for heat of formation

1 mole of product should be formed

also

the reactants should be elements in their standard state

A) oxygen should be in 02 form

B) hydrogen should be in H2 form

C) 1 mole of product should be formed

so

the correct answer is

d) 0.5 N2 + 02 --> N02

35) we know that

PM = dRT

so

1 x M = 5.75 x 0.0821 x 273

M = 129 g/mol

so A) 129

36) c) a gas assumes the volume of its container

37)

moles = mass / molar mass

moles of KCl03 = 7.5 / 123 = 0.061

now

moles of 02 formed = 1.5 x moles of KCl03

moles of 02 = 1.5 x 0.061 = 0.0915

now

PV = nRT

1 x V = 0.0915 x 0.0821 x 298

V = 2.2 L

so

the answer is D) 2.2

38) a negative dH corresponds to exothermic reaction

a positive dH corresponds to endothermic reaction

so

answer is B) negative ,exothermic

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