Acommon laboratory reaction is the neutralization of an acid with a base. when 50.0 ml of 0.500 m hcl at 25.0°c is added to 50.0 ml of 0.500 m naoh at 25.0°c in a coffee cup calorimeter, the temperature of the mixture rises to 28.2°c. assume the mixture has a specific heat capacity of 4.18 j/(g. k) and the densities of all solutions are 1.00 g/ml

a. calculate the qrxn
b. calculate deltah rxn per mole of acid
c. when this reaction is repeated with 50.0 ml of 2.0 m hci at 25.0°c, the temperature of the mixture also increases to 28.2°c. explain

a) 1.34 x 10 ³ j

b) 5.35 x 10⁴ j

c) See explanation below.

Explanation:

a) q rxn = m mix x C x ΔT = ( 100 mL x 1.00 g/mL) 4.18 J/g/K x (3.2 K)  =

1.34  x 10³ j

b) mol HCl = (50mL x 1L/1000 mL) x 0.500 mol/L = 0.0025 mol

1.33 x 10³ j/ 0.0025 mol x 1mol =  5.35 x 10⁴ joule

c) The heat of  reaction is heat released in the neutralization, if we change the amount of HCl to neutralize but do not increase the equivalent number of moles of NaOH, the heat released will be the same.

i think it is d. e because the next one is igneous.

if you mean hydrogen and oxygen then water is the product