Answer is: concentration of products increases (ammonia nad water).
Chemical reaction: heat + NH₄⁺ + OH⁻ ⇄ NH₃ + H₂<span>O.
</span>According to Le
Chatelier's Principle, the position of equilibrium moves to counteract the
change, because heat is increased, system consume that heat, so equilibrium is shifted to right, by decreasing concentration of reaactants and increasing concentration of product.
Answer:
Below:
Explanation:
To calculate an energy change for a reaction: add together the bond energies for all the bonds in the reactants - this is the 'energy in' add together the bond energies for all the bonds in the products - this is the 'energy out.
Hope it helps....
It's Muska
Answer:
A - The can was open and gases were released.
<u>Answer:</u>
<u>For a:</u> The edge length of the unit cell is 314 pm
<u>For b:</u> The radius of the molybdenum atom is 135.9 pm
<u>Explanation:</u>
To calculate the edge length for given density of metal, we use the equation:
where,
= density = 
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
= Avogadro's number = 
a = edge length of unit cell =?
Putting values in above equation, we get:
![10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm](https://tex.z-dn.net/?f=10.28%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%20%28a%29%5E3%7D%5C%5C%5C%5Ca%5E3%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%2010.28%7D%3D3.099%5Ctimes%2010%5E%7B-23%7D%5C%5C%5C%5Ca%3D%5Csqrt%5B3%5D%7B3.099%5Ctimes%2010%5E%7B-23%7D%7D%3D3.14%5Ctimes%2010%5E%7B-8%7Dcm%3D314pm)
Conversion factor used: 
Hence, the edge length of the unit cell is 314 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:
where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:
Hence, the radius of the molybdenum atom is 135.9 pm
The question requires us to explain the differences in radii of neutral atoms, cations and anions.
To answer this question, we need to keep in mind that a neutral atom presents the same number of protons (positive particles) and electrons (negative particles). Another important information is that the protons are located in the nucleus of the atom, while the electrons are around the nucleus. Also, there is an electrostatic force between protons and electrons, which means that they the protons tend to attract the electrons to the nucleus.
While a neutral atom presents the same number of protons and electrons, a cation is an ion with positive charge, which means it has lost one or more electrons. In a cation, the balance between protons and electrons doesn't exist anymore: now, there is more positive than negative charge (more protons than electrons), and the overall attractive force that the protons have for the electrons is increased. As a result, the electrons stay closer to the nucleus and the radius of a cation is smaller than the neutral atom from which it was derived.
On the other side, anions present negative charge, which means they have received electrons. Similarly to cations, the balance between protons and electrons doesn't exist anymore, but in this case, there are more electrons than protons. In an anion, the overall attractive force that the protons have for the electrons is decreased. As a result, the electrons are "more free" to move and, as they are not so attracted to the nucleus, they tend to stay farther from the positive nucleus compared to the neutral atom - because of this, the radius of an anion is larger than the neutral atom from which it was derived.
