Which equation represents neutralization? 6Li(s) + N2(g) → 2Li3N(S)
2Mg(s) + O2(g) → 2MgO(S)
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)
Pb(NO3)2(aq) + K2CrO4(aq) → 2KNO3(aq) PbCrO4(s)

2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

Explanation:

A neutralization reaction refers to any reaction between an acid and a base to form salt and water only. In a neutralization reaction, the key reaction that takes place to produce water is shown ionically below;

H^+(aq) +OH^-(aq) > H2O(l)

Looking at all the options provided, our gaze will have to settle on the reaction;

2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

We can clearly see that there is an acid that reacts with a base to form a salt and water. The base contains the OH^- ion while the acid contains the hydrogen ion.

This is the only reaction of this sort among the options hence it is the correct answer.

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol