There is 67 gram of CO2 produced in the Octane reaction which is come after 2 SF
Answer:
E = 2.05×10⁻²³ J
Explanation:
Given data:
Wavelength of transmission = 9.7 mm (9.7/1000 = 0.0097 m)
Energy of transition = ?
Solution:
9.7 mm (9.7/1000 = 0.0097 m)
9.7×10⁻³ m
Formula:
E = hc/λ
By putting values,
E = 6.63 ×10⁻³⁴ Js × 3×10⁸ m/s / 9.7×10⁻³ m
E = 19.89×10⁻²⁶ J.m / 9.7×10⁻³ m
E = 2.05×10⁻²³ J
Answer is: boiling point will be changed by 4°C.
Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).
Change in boiling point: ΔT =i · Kb · b.
Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.
b - molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.
i - Van't Hoff factor.
Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.
Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.
Answer:
Answer: V=67.2 L
Explanation:
Ideal Gas Law: PV=nRT
P=1.00 atm (STP)
V=?
n=3.00 mol
R=0.08206Latm/Kmol
T=273.15 K (STP)
To find V, we would manipulate the equation to V=nRT/P
With significan figures, our answer is V=67.2 L.
Answer:
P.E = 25.48 J
Explanation:
Given data:
Mass = 2 Kg
Height = 1.3 m
Potential energy = ?
Solution:
Formula:
P.E = m . g . h
P. E = potential energy
m = mass in kilogram
g = acceleration due to gravity
h = height
Now we will put the values in formula.
P.E = m . g . h
P.E = 2 Kg . 9.8 m /s² . 1.3 m
P.E = 25.48 Kg. m² / s²
Kg. m² / s² = J
P.E = 25.48 J
