What are the smallest whole number coefficients the reactants and product when t

Question

What are the smallest whole number coefficients the reactants and product when the following molecular equation is correctly balanced?_nh_3(g) +_O_2(g) rightarrow_no(g) +_h_2O(g) 1, 1, 1, 1, 2, 3, 2, 3 3, 2, 3, 4 4, 5, 4, 6 At constant pressure and 25degreeC what is delta_tHdegree for the following reaction If the complete consumption of 14 5 g of C_2H_6 liberates 752.3 kj of heat energy 3120 kJ/mol-rxn 1560 kj/md-rxn 725 kJ/mol-rxn 362 kJ/mol-rxn 788.41 kJ/mol-rxn A 170.0-g sample of metal at 73.00degreeC is added to 170 0 g of H_2O(l) at 26.00degreeC in an insulated container. The temperature rises to 27.70degreeC Neglecting the heat capacity of the container, what is the specific heat capacity of the metal? The specific heat capacity of H_2O(t) is 4 18 J/(g degreeC). 4.18 J/(g degreeC) 111 J/(g degreeC) 0.157 J/(g degreeC) 0.157 J/(g degreeC) 26.6J/(gdegreeC) Specific heat capacity is the quantity of heat needed to change the temperature of 1.00 g of a substance by 1 K. the quantity of heat needed to change the temperature of 1.00 g of a substance by 4.184 K. the capacity of a substance to gain of lose a 1.00 J of energy in the form of heat. the temperature change undergone when 1.00 g of a substance absorbs 4.184 J. the maximum amount of energy in the form of heat that 1.00 g of a substance may absorb without decomposing.

Explanation / Answer

13) Let us write the balanced equation first

4NH3 + 5O2 -->4NO + 6H2O

Total hydrogen on both side = 12

Total nitrogen on both side = 4

total oxygen on both side = 10

So answer is 4,5,4,6

14) 14.5 g of C2H6 liberates =752.3 KJ of energy

So 1 gram will liberate = 752.3 / 14.5 KJ of energy

so 30 gram (1mole) will liberate = 30 X 752.3 / 14.5 KJ = 1556.48 KJ / mole = 1560 KJ / mole

As per given equation two moles are undergoing combustion so heat liberated will be =2 X 1556.48 = 3112.96 KJ / per rxn

So correct answer rounded to near digits = 3120 KJ / rxn

15) Heat absorbed = Heat lose

Heat absorbed by water = Mass of water X specific heat of water X (change in temperature)

Heat abosorbed by water = 170 X (4.18) ( 27.7-26) = 1208.02 Joules

Heat lost by metal = Mass of metal X specific heat of metal X (change in temperature)

1208.02 = 170 X specific heat of metal X (73-27.7)

specific heat of metal = 0.1568 Or 0.157 J / g 0C

16) The heat needed to change the temperature of 1 g of substance by 1K or 10C

As per given equation

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