n = number of moles of sample of octane = 0.07 mol
Q = energy absorbed by a sample of octane = 3.5 x 10³ J
c = molar heat capacity of octane = 254.0 J/K* mol
ΔT = increase in temperature of octane = ?
Heat absorbed is given as
Q = n c ΔT
inserting the values
3.5 x 10³ J = (0.07 mol) (254.0 J/K* mol) ΔT
ΔT = (3.5 x 10³ )/((0.07) (254.0))
ΔT = 196.85 K
hence increase in temperature comes out to be 196.85 K
Answer:
Your answer would be Ionic bond.
Answer:
55.18 L
Explanation:
First we convert 113.4 g of NO₂ into moles, using its molar mass:
- 113.4 g ÷ 46 g/mol = 2.465 mol
Then we<u> use the PV=nRT formula</u>, where:
- P = 1atm & T = 273K (This means STP)
- R = 0.082 atm·L·mol⁻¹·K⁻¹
Input the data:
- 1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K
And <u>solve for V</u>:
Water is always the solvent