Answer:elements cannot be split into two
Explanation:elements are pure substances which cannot be split into two by chemical means
Answer:
Explanation:
To start with, the mass of a substance is centered on the amount of particles contained in it.
Mass of a substance measure is the same anywhere it is taken.
In chemistry, the concept of mole is used to find the exact estimate of the number of entities in a sample.
A mole of substance is quantity that is used to express the number of particles contained in it.
1 mole of any substance contains the Avogadro's number of particles.
This amount of particles is given as 6.02 x 10²³; the particles can be protons, neutrons, electrons e.t.c.
To find the number of moles;
Number of moles = 
Through this, we clearly ascertain the number of entities in a sample given the mass.
Answer:
Explanation:
Un ácido de Lewis es una especie química que contiene un orbital vacío que es capaz de aceptar un par de electrones de una base de Lewis para formar un aducto de Lewis
<u>Answer:</u> The amount of energy released per gram of
is -71.92 kJ
<u>Explanation:</u>
For the given chemical reaction:

The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28reactant%29%7D%5D)
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
Taking the standard enthalpy of formation:

Putting values in above equation, we get:
![\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%285%5Ctimes%20%281271.94%29%29%2B%289%5Ctimes%20%28-285.83%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_%7Brxn%7D%3D-9078.57kJ)
We know that:
Molar mass of pentaborane -9 = 63.12 g/mol
By Stoichiometry of the reaction:
If 2 moles of
produces -9078.57 kJ of energy.
Or,
If
of
produces -9078.57 kJ of energy
Then, 1 gram of
will produce =
of energy.
Hence, the amount of energy released per gram of
is -71.92 kJ
A mixture, because it contains multiple dyes and compounds
hopefully i could help ;)