<em><u>Given</u></em>
Weight of pottasium fluoride (KF) = 29 gms
<em><u>To find</u></em>
Number of moles of pottasium fluoride = ?
<em><u>Solution</u></em>
we know that molar mass of KF = 58.0967 g
Now,

<em>Number of moles = 29/58 = 0.5 moles</em>
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M is the unit for concentration (C)
The units for concentration is mol/ L
To find moles you would have to multiply concentration x volume to cancel out the volume and find get moles (n)
For ex:
Moles (n) = C x V (concentration x volume)
N = 1.23 mol/L x 5L (vol. gets cancelled out)
N = 6.15 moles (you get left with moles)
OR to find concentration:
C = 6.15 moles / 5L
C = 1.23 mols/L OR 1.23 M
Answer:
The precipitate will form.
Explanation:
Let's write the equilibrium expression for the solubility product of calcium sulfate:
⇄ 
The solubility product is defined as the product of the free ions raised to the power of their coefficients, in this case:
![K_{sp}=[Ca^{2+}][SO_4^{2-}]=10^{-4.5}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D10%5E%7B-4.5%7D)
Our idea is to find the solubility quotient, Q, and compare it to the K value. A precipitate will only form if Q > K. If Q < K, the precipitate won't form. In this case:
![Q_{sp}=[Ca^{2+}][SO_4^{2-}]=5.00\cdot10^{-2} M\cdot7.00\cdot10^{-3} M=3.5\cdot10^{-4}](https://tex.z-dn.net/?f=Q_%7Bsp%7D%3D%5BCa%5E%7B2%2B%7D%5D%5BSO_4%5E%7B2-%7D%5D%3D5.00%5Ccdot10%5E%7B-2%7D%20M%5Ccdot7.00%5Ccdot10%5E%7B-3%7D%20M%3D3.5%5Ccdot10%5E%7B-4%7D)
Now given the K value of:

Notice that:

This means the precipitate will form, as we have an excess of free ions and the equilibrium will shift towards the formation of a precipitate to decrease the amount of free ions.
Answer:
<em>Different rocks have high, medium, and low melting points.</em>
Explanation:
There is a considerable range of melting temperatures for different compositions of magma. All the silicates are molten at about 1200°C (when a part of rock) and all are solid when cooled to about 600°C. Often the silicates are grouped as high, medium and low-melting point solids.
Answer:

Explanation:
Let's firstly identify the atomic number (the number of protons) of Pu. This is done by referring to the periodic table and finding Pu. The atomic number of Pu is:

In order to identify the type of a nuclear decay, we need to find the N/Z ratio. This is the ratio between the number of neutrons and the atomic number of an isotope. The number of neutrons is found by subtracting the number of protons from the mass number:

That said, the N/Z ratio equation becomes:

This is a relatively high number thinking about the belt of stability of isotopes. Ideally, stable isotopes with a low Z value have an N/Z ratio of 1. Heavier isotopes with Z > 50 would have a slightly higher N/Z ratio and would be stable around N/Z = 1.25. This means we wish to decrease the N/Z ratio as much as possible.
Among all the decays, alpha-decay is preferred to decrease the N/Z ratio significantly (1.45 is much higher than 1.25). That said, we'll release an alpha particle with some nucleotide X of mass M and atomic number Z:

According to the mass and charge conservation law:


Identify an element with Z = 92 in the periodic table. This is uranium, U:
