URGENT A 2.00 g marshmallow is combusted under a flask containing 50.0 g of water at an initial temperature of 25.0°C. After the combustion of the marshmallow, 0.30g of charred marshmallow remained and the temperature of the water increased to 75.0°C. If the energy associated by the combustion of the marshmallow is 2.5 kilocalories. What is the percent error if the packaging label says that there are 3.5 kilocalories per gram of marshmallow.

Specific heat of water is 4.184 J/g °C or 1 cal/g°C

Percent Error = 100 x (Accepted value - Experimental value)/Accepted value

A. 42% error

B. 58% error

C. 64% error

D. 29% error

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

the formula mass of a molecule or a compound is the sum of the atomic weights of the atoms in its empirical formula

(that's what i know about formula mass)

може ли да зареждате повече информация, моля