Mass of Magnesium bromide : 5.76 g
<h3>Further explanation</h3>
Complete question
<em>What mass of Magnesium bromide is formed when 1.00 g of magnesium reacts with 5.00 g of bromine?</em>
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Reaction
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mol Mg

mol Br₂

Limiting reactants : Br₂(smaller)
mol MgBr₂ = mol Br₂=0.0313
mass MgBr₂ :

Answer:
See explanation.
Explanation:
1. There are 8 electrons. Elements that end with 2p orbitals are in the 2nd period (aka row) of the periodic table. Elements that have 4 electrons in 2p are in the 16th group (aka column) (column 16 may also be referred to as 6A) of the periodic table. So looking at row 2, column 16, we can see that the first diagram is of O, Oxygen.
2. 8 electrons. This is the same diagram as the one above.
3. 13 electrons. Elements ending with 3p are in period 3. Elements with 1 valence electron in a p orbital are in group 13 (aka group 3A).
4. 7 electrons. We already know 2p is period 2. 3 valence electrons in a p orbital means that it is in group 15/group 5A.
I did not write the answers for #3 and 4 but they can be easily found on a periodic table with the info I gave.
Answer:
6.6666666667 feet per second I believe
Answer:
The maximum amount of work that can be done by this system is -2.71 kJ/mol
Explanation:
Maximum amount of work denoted change in gibbs free energy
during the reaction.
Equilibrium concentration of B = 0.357 M
So equilibrium concentration of A = (1-0.357) M = 0.643 M
So equilibrium constant at 253 K, ![K_{eq}= \frac{[B]}{[A]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%20%5Cfrac%7B%5BB%5D%7D%7B%5BA%5D%7D)
[A] and [B] represent equilibrium concentrations

When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M
So reaction quotient at this situation, 
We know, 
where R is gas constant and T is temperature in kelvin
Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and
is 0.555
So, 
= -2710 J/mol
= -2.71 kJ/mol