Suppose you place 0.0600 g of magnesium chips in a coffee-cup calorimeter and then add 100.0 mL of 1.00 M HCl. The reaction that occurs is Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq) The temperature of the solution increases from 23.08 °C (296.23 K) to 25.78 °C (298.93 K). What is the enthalpy change for the reaction per mole of Mg? Assume a specific heat capacity of the solution is 4.20 J/g∙K and the density of the HCl solution is 1.00 g/mL. 459.55 kJ/mol Show ApproachHide Tutor Steps TUTOR STEP What is the heat released by the reaction, qr? qr + qsolution = 0 qsolution = 1130 J

ΔH = -459.1 kJ/mol

Explanation:

Step 1: Data given

Mass of the magnesium chips = 0.0600 grams

Volume of a 1.00 M HCL = 100.0 mL= 0.100 L

The temperature of the solution increases from 23.08 °C (296.23 K) to 25.78 °C (298.93 K)

The specific heat capacity of the solution is 4.20 J/g∙K

density of the HCl solution is 1.00 g/mL

Step 2: The balanced equation

Mg(s) + 2 HCl(aq) → H2(g) + MgCl2(aq)

Step 3: Calculate heat

Q = m*C*ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with m = the mass 100 mL * 1g/mL = 100 grams

⇒with C = the specific heat capacity of the solution is 4.20 J/g*K

⇒with ΔT = the change of temperature = T2 - T1 = 298.93 - 296.23 = 2.7 K

Q = 100 * 4.20 * 2.7

Q = 1134 J

Step 4: Calculate moles Mg

Moles Mg = mass Mg / molar mass Mg

Moles Mg = 0.0600 grams / 24.305 g/mol

Moles Mg = 0.00247 moles

Step 5: Calculate  the enthalpy change for the reaction per mole of Mg

ΔH = -1134 J / 0.00247 moles

ΔH = -459109 J/mol = -459.1 kJ/mol

(negative since it's an exothermic reaction)

m=9

step-by-step explanation:

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.