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.
The answer is 1.98 × 10^22 iron (II) ions
Answer:
25.0g is the mass of sulfur
Explanation:
The sulfur and the oxygen are 50:50. That means there is 1 mole of S per mole of O.
To solve this question we need to convert the mass of O to moles, as moles of O = Moles of S, we can find the moles of S and its mass:
<em>Moles O = Moles S:</em>
12.5g O * (1mol / 16g) = 0.781 moles
<em>Mass S:</em>
0.781 moles * (32g / mol) =
<h3>25.0g is the mass of sulfur</h3>
<span>3.68 liters
First, determine the number of moles of butane you have. Start with the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass butane = 4*12.0107 + 10*1.00794 = 58.1222 g/mol
Moles butane = 2.20 g / 58.1222 g/mol = 0.037851286
Looking at the balanced equation for the reaction which is
2 C4H10(g)+13 O2(g)→8 CO2(g)+10 H2O(l)
It indicates that for every 2 moles of butane used, 8 moles of carbon dioxide is produced. Simplified, for each mole of butane, 4 moles of CO2 are produced. So let's calculate how many moles of CO2 we have:
0.037851286 mol * 4 = 0.151405143 mol
The ideal gas law is
PV = nRT
where
P = Pressure
V = Volume
n = number of moles
R = Ideal gas constant ( 0.082057338 L*atm/(K*mol) )
T = absolute temperature (23C + 273.15K = 296.15K)
So let's solve the formula for V and the calculate using known values:
PV = nRT
V = nRT/P
V = (0.151405143 mol) (0.082057338 L*atm/(K*mol))(296.15K)/(1 atm)
V = (3.679338871 L*atm)/(1 atm)
V = 3.679338871 L
So the volume of CO2 produced will occupy 3.68 liters.</span>
Answer:
The answer is below
Explanation:
Here some of the difference between aluminum phosphide and aluminum phosphate
1. Aluminum phosphide is a crystalline solid with either dark gray or yellow color. While Aluminum Phosphate is in a nanopowder form but with a whitish color.
2. Aluminum phosphide is used in the fumigation of plants, while Aluminum phosphate is used in cosmetics, paints, etc.
3. Aluminum phosphide has a compound formula of AlP with a molar mass of 57.9552 g/mol, while Aluminum Phosphate has a compound formula of AlO4P with a molar mass of 121.9529 g/mol
