How much heat is released by the burning of a nut that causes a cup containing 50g of water to experience an increase in temperature from 25 degrees celsius to 35 degrees celsius

2093 J is released by the burning of the nut

Explanation:

Calorimetry formula:  Q = m.C.ΔT

Q = 50g . C .ΔT

C is the specific heat of water: 4.186 J/g°C

ΔT = Final temperature - Initial temperature

ΔT = 35° - 25°  =  10°C

Q = 50g . 4.186 J/g°C . 10°C → 2093 joules

There is 2092 J of heat released

Explanation:

Step 1: Data given

Mass of the water = 50.0 grams

Initial temperature = 25.0 °C

Final temperature = 35.0 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the heat released

Q = m*c*ΔT

⇒ m = the mass of water = 50.0 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change of temperature = T2 - T1 =  35.0 - 25.0 = 10.0 °C

Q = 50.0g * 4.184 J/g°C * 10.0°C

Q = 2092 J

There is 2092 J of heat released

i believe it is 40–40–20%

explanation:

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 :