A tissue sample at 275 K is submerged in 2 kg of liquid nitrogen at 70 K for cryopreservation. The final temperature of the nitrogen is 75 K. What is the heat capacity of the sample in J/K

heat capacity of the sample = 37.8 J/K

Explanation:

Step 1: Data given

Temperature of the sample = 275 K

The mass of liquid nitrogen = 2kg

temperature of liquid nitrogen = 70 K

The final temperature of the nitrogen is 75 K

Step 2: Calculate heat

Q = m*c*ΔT

⇒with m = the mass of liquid nitrogen = 2 kg = 2000 grams

⇒with c= the specific heat of the liquid nitrogen = 1.04 J/g*K

⇒with ΔT = the change of temperature of liquid nitrogen = T2 - T1 = 75 - 70 = 5K

Q = 2000 grams * 1.04 J/g*K * 5K

Q = 10400 J

Step 3: Calculate the heat capacity of the sample

heat capacity of the sample = 10400 J / 275 K

heat capacity of the sample = 37.8 J/K

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :