Answer: A 59.5 degree celcius
The equation that we will use to solve this problem is :
PV = nRT where:
P is the pressure of gas = 1.8 atm
V is the volume of gas = 18.2 liters
n is the number of moles of gas = 1.2 moles
R is the gas constant = 0.0821
T is the temperature required (calculated in kelvin)
Using these values to substitute in the equation, we find that:
(1.8)(18.2) = (1.2)(0.0821)(T)
T = 332.5 degree kelvin
The last step is to convert the degree kelvin into degree celcius:
T = 332.5 - 273 = 59.5 degree celcius
Answer:
Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:
We have 
as a reactive in the target reaction and is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A:
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.
C=10⁻⁶ mol/L
pH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg10⁻⁶=14-6=8
B. pH = 8
It would be a positive charge because it lost two electrons, if the charge was neutral it would be the same amount of protons and electrons, if the charge was negative the electrons would be 20 instead of 18. So in this case it is positive.
Answer:
[Ne]3s2
Explanation:
ahora tenemos que mirar cada una de las configuraciones electrónicas de cada átomo de cerca antes de tomar una decisión.
considerando la configuración electrónica más externa de cada una de las especies mostradas;
para la primera configuración, ns2 np6 corresponde a un gas noble.
para la segunda configuración ns2 np3 corresponde a un elemento no metálico del grupo 5.
para la tercera configuración, ns2 corresponde a un elemento metálico del grupo 2.
para la cuarta configuración, ns2 np4 corresponde a un elemento no metálico del grupo 6
