Astudent mixes a 10.0 ml sample of 1.0 m naoh with a 10.0 ml sample of 1.0 m hcl in a polystyrene container. the temperature of the solutions before mixing was 20 c.

if the final temperature of the mixture is 26 c, what is the experimental value of delta h?

assume the solution mixture has a specific heat of 4.2 j/gc and a density of 1 g/ml.a. -50 kj/molb. -25 kj/molc. -50,000 kj/mold. -500 kj/mol

The experimental value of ΔH is -50 kJ/mol

Explanation:

Step 1: Data given

Volume of 1.0 M NaOH = 10.0 mL = 0.01 L

Volume of 1.0 M HCl = 10.0 mL = 0.01 L

Temperature before mixing = 20 °C

Final temperature = 26 °C

Specific heat of solution = 4.2 J/g°C

Density = 1g/mL

Step 2: Calculate q

q = m*c*ΔT

⇒ with m = the mass

  ⇒ 20.0 mL * 1g/mL = 20 grams

⇒  c = specific heat of solution = 4.2 J/g°C

⇒ ΔT = T2 -T1 = 26 -20 = 6 °C

q = 20g * 4.2 J/g°C * 6°C

q = 504 J

ΔHrxn = -q  ( because it's an exothermic reaction)

ΔHrxn = -504 J

Step 3: Calculate number of moles

Moles = Molarity * volume

Moles = 1M *0.01 L = 0.01 moles

Step 4: Calculate the experimental value of ΔH

ΔHrxn = -504 / 0.01 mol = -50400 J/mol = -50.4 kJ/mol

The experimental value of ΔH is -50 kJ/mol

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Explanation:2+2=4-1=3 quick math