Answer:
0.56L
Explanation:
This question requires the Ideal Gas Law:
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.
Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:
Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means
and 
Lastly, we must calculate the number of moles of
there are. Given 0.80g of
, we will need to convert with the molar mass of
. Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2: 
Thus, 
Isolating V in the Ideal Gas Law:


...substituting the known values, and simplifying...


So, 0.80g of
would occupy 0.56L at STP.
Answer:
Option D is good to go!
Explanation: as per the reactivity series more reactive substances will react with the counterpart substance.The most reactive substance here is calcium while the least reactive is aluminium, the magnesium comes in between.As per their reactivity, these substances will react with oxygen.
Explanation:
Answer: 8.830418848725065
Explanation:
8.830418848725065
I think Both protons and neutrons (and their anti-particles) froze out at 1013 K, about 0.0001 seconds after the Big Bang. Protons and neutrons are sub atomic particles of an atom that are found in the nucleus of an atom. Proton is the positively charge particle while the neutron has no charge. The proton positive charge accounts for the positive nuclear charge.
Answer:
a)
b)
Explanation:
a) The reaction:

The free-energy expression:

![E=E_{red}-E_{ox]](https://tex.z-dn.net/?f=E%3DE_%7Bred%7D-E_%7Box%5D)
The element wich is reduced is the Fe and the one that oxidates is the Mg:

The electrons transfered (n) in this reaction are 2, so:


b) If you have values of enthalpy and enthropy you can calculate the free-energy by:

with T in Kelvin

