A 1000-g iron frying pan is placed on a stove. The pan increases from 20°C to 130°C. If the specific heat of iron is 0.444 J/goC, how much heat was gained by the pan? A. -48,840 J
B. 48,840 J
C. 57,720 J
D. - 57,720 J

48,840 J

Explanation:

First of all, you can use process of elimination to take out the negatives, since you can't gain negative heat (in this situation).

Now just for the Math:

Remember this equation — Q = mcΔt

The variables for the equation are:

— m - The measure in grams or kilograms.

— c - The measure of the specific heat.

— Δt - The measure the change in temperature.

Now find these variables:

— m = 1000 g

— c = 0.444 J/goC

— Δt = 130 - 20 = 110 C

Now plug these numbers into the equation and solve!

Q = mcΔt

If you multiply the number you would get 48,840 J. You also have to cancel out the units as well.

Hope it Helped!

Pls Give Brainliest!

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 :

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be