Answer:
For many solids dissolved in liquid water, the solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.
Explanation:
We are given with
136 g P4
excess oxygen
The complete combustion reaction is
P4 + 5O2 => 2P2O5
Converting the amount of P4 to moles
136/123.9 = 1.098 moles
Using stoichiometry
moles P2O5 = 1.098 x 2 = 2.195 moles P2O5
1.123 nano-grams is your answer, do you understand now gimme dat 5 star and brainiest
Answer:
The answer is
<h2>19.31 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula

From the question
mass of gold = 475.09g
volume = 24.6 cm³
The density of the gold is

We have the final answer as
<h3>19.31 g/cm³</h3>
Hope this helps you
The balanced chemical equation for the combustion of butane is:

Δ
= Σ
Δ
-Σ
Δ
=
=[-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