If a system absorbs 300.5J of heat and performs 1.50kJ of work on the surroundings, what is the ΔE?

Complete question:

If a system absorbs 300.5J of heat and performs 1.50kJ of work on the surroundings, what is the change in internal energy of the system ΔE?

Change in internal energy of the system is 1800.5 J

Explanation:

Given:

heat transferred to the system, Q = 300.5J

work done on the surroundings, W = 1.50kJ

Change in internal energy of the system ΔE, can be calculated by applying thermodynamic equation of change in internal energy, work done and heat transferred.

ΔE = Q  + W

Where;

ΔE is change in internal energy

Q is heat transfer

W is work done

ΔE = 300.5 J  +  1500 J

ΔE = 1800.5 J

Therefore, change in internal energy of the system is 1800.5 J

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

nah.   i'm just sleepy.