Answer:
205 K (to 3 significant figures)
Explanation:
Assuming that 4 moles of the gas behaves like an ideal gas and obey the kinetic molecular theory.
Let's apply the ideal gas law, pV= nRT.
Here p denotes the pressure of the gas, V is for volume, n is the number of moles of the gas, R is the universal gas constant and T is the temperature.
Substitute the given information into the equation:
5.6 atm ×12 L= 4 mol ×R ×T
Since pressure is in atm and volume is in L, we can use R= 0.08206 L atm K⁻¹ mol⁻¹.
5.6 atm ×12 L= 4 mol ×0.08206 L atm K⁻¹ mol⁻¹ ×T
T= 67.2 ÷0.32824
T= 204.73 (5 s.f.)
T= 205 K (3 s.f.)
Answer:
Its mass is measured on a scale.
It is measured on a balance.
Explanation:
The mass of an object can easily be determined using a balance or weighing scale.
Mass is the amount of matter contained in a substance. it is has the same value every where and will not vary by geography or location.
Weight is the force on body due to gravity. It is a function of mass and acceleration due to gravity.
Both mass and weight are different.
They can both be measured using a weight device.
Answer:
2.7 g/mL:)
An aluminum bar was found to have a mass of 27g. Using water displacement, the volume was measured to be 10 ml. What is the density of the aluminum? Group of answer choices (27 g)/(10 ml) (10 ml )/(2.70 g) (270 g)/(10 ml) (10 ml )/(27 g)
Answer:
When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is not affected.
Explanation:
The equilibrium constant of a reaction is known to remain steady.
Even if all the coefficients of all the species in the reaction are multiplied by two, the value of the equilibrium constant will reamin the same because the equilibrium position will not change as a result of that.
The balanced chemical equation for the combustion of butane is:
Δ
= Σ
Δ
-Σ
Δ
= ![[{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]](https://tex.z-dn.net/?f=%5B%7B8%2A%28-393.5kJ%2Fmol%29%7D%2B%7B10%2A%28-241.82kJ%2Fmol%29%7D%5D-%5B%7B2%2A%28-125.6kJ%2Fmol%29%7D%2B13%2A%280%20kJ%2Fmol%29%7D%5D)
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:
Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol
