How much heat is produced when 100 ml of 0.250 m hcl (density, 1.00 g/ml) and 200 ml of 0.150 m naoh (density, 1.00 g/ml) are mixed? +⟶+δh°=−58kj if both solutions are at the same temperature and the heat capacity of the products is 4.19 j/g °c, how much will the temperature increase? what assumption did you make in your calculation?

1.45 kilo joules of heat is produced when 100 mL of 0.250 M HCl and 200 mL of 0.150 M NaOH.

1.15°C is the temperature increase.

Step-by-step explanation:

,ΔH°=-58 kJ/mol

Molarity of HCl = 0.250 M

Volume of HCl = 100 ml = 0.1 L

Moles of HCl = n

Molarity of NaOH= 0.150 M

Volume of NaOH= 200 ml = 0.2 L

Moles of NaOH= n'

According to reaction, 1 mol of HCl reacts with 1 mol of NaOH. Then 0.0250 mole of HCl will reacts with 0.0250 mol of NaOH.

of NaOH

As we can see that moles of NaOH are in excess.Hence, excessive agent.

The enthalpy of the reaction = ΔH°=-58 kJ/mol

Energy released when 0.0250 moles of HCl reacted with 0.0250 moles of NaOH:

(Negative sign indicates that heat is liberated.)

Mass of the HCL solution = m

Volume of HCl ,v= 100 ml

Density of HCl solution = d = 1.00 g/mL

Mass of NaOH solution = m'

Volume of NaOH ,v' = 200 ml

Density of NaOH solution = d' = 1.00 g/mL

Mass of the solution after mixing,M = m + m' = 100  g + 200 g = 300 g

Heat absorbed by the final solution formed after mixing = Q'

Heat absorbed by the final solution formed after mixing = Heat released during reaction

Q' = -Q = -(-1.45 kJ)= 1.45 kJ=1450 J  (1kJ = 1000 J)

Q' = mcΔT

generally, m = mass of the substance

c  = specific heat of the substance

ΔT = Change in temperature

Specific heat capacity of the product formed after mixing = c = 4.19 J/g°C

Mass of the resulting  mix = 300 g

ΔT = ?

1.15°C is the temperature increase.

huh? im pretty sure it is in 199

step-by-step explanation:

well my understanding of this question is they want you to use the y=14.1x -2.3 and put the coordinates they gave you onto the line segment of   y=14.1x -2.3

step-by-step explanation: