Answer:
La respuesta está en la explicación
Explanation:
Los átomos que no poseen ___<em>CARGA</em>____ son conocidos como ____<em>ÁTOMOS</em>____ ____<em>NEUTROS</em>_____. Los átomos que poseen carga se denominan ___<em>IONES</em>____, aquellos con carga positiva de llaman ____<em>PROTONES</em>____ y son aquellos que __<em>PIERDEN</em>__ electrones. Los ______<em>ANIONES</em>____ son aquellos con carga negativa y son los que _____<em>GANAN</em>___ electrones. Para todos los casos el valor de _<em>MASA</em>__ y _<em>NÚMERO ATÓMICO</em>_ permanecen iguales y son los que sacamos de la tabla periódica.
So C12 is the limiting reactant and P4 is the excess
Mass P4 consumed= 0.31 mol X 123.9 g/mol =38.41 g P4 consumed.
Hope i helped
Answer:
A) Has properties of both metals and nonmetals - Barium.
B) Nonreactive gas - Neon.
C) Great conductor of heat and electricity - Boron.
D) Malleable and highly reactive - Potassium.
Explanation:
hope it helps .
Answer:
- From octane: 
- From ethanol: 
Explanation:
Hello,
At first, for the combustion of octane, the following chemical reaction is carried out:

Thus, the produced mass of carbon dioxide is:

Now, for ethanol:


Best regards.
Answer:
Final temperature = 1279.25 K
Explanation:
We can solve this using the formula for Charles law since we are given volume and temperature.
From Charles law, we know that;
V1/T1 = V2/T2
Where;
T1 is the initial temperature
V1 is the initial volume
T2 is the final temperature
V2 is the final volume
We are given;
V1 = 2 L
T1 = 301 K
V2 = 8.5 L
Thus, making T2 the subject, we have;
T2 = V2•T1/V1
Plugging in the relevant values;
T2 = 8.5 × 301/2
T2 = 1279.25 K