The element with 4 protons in the nucleus in Beryllium.
Answer: I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
Explanation:
An Oxidant is any substance that oxidizes, or receives electrons from, another; in so doing, it becomes reduced in oxidation number.
A Reductant thus exactly the opposite.
Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:
I2 --> 2I-.
The oxidation number reduced from 0 to -1.
In contrast, the oxidation number of 2S2O3(-2) increases from -4 to -2.
Thus, I2 is the Oxidant; while the 2S2O3(-2) is the reductant.
A hydrate is a chemical that has water molecules loosely bonded to it. The water molecules are not ... You will be using the hydrate CuSO4 . ?H2O. Sample Calculation-. An empty crucible has a mass of 12.770 grams.
Answer:
Density, melting point. and magnetic properties
Explanation:
I can think of three ways.
1. Density
The density of Cu₂S is 5.6 g/cm³; that of CuS is 4.76 g/cm³.
It should be possible to distinguish these even with high school equipment.
2. Melting point
Cu₂S melts at 1130 °C (yellowish-red); CuS decomposes at 500 °C (faint red).
A Bunsen burner can easily reach these temperatures.
3. Magnetic properties
You can use a Gouy balance to measure the magnetic susceptibilities.
In Cu₂S the Cu⁺ ion has a d¹⁰ electron configuration, so all the electrons are paired and the solid is diamagnetic.
In CuS the Cu²⁺ ion has a d⁹ electron configuration, so all there is an unpaired electron and the solid is paramagnetic.
A sample of Cu₂S will be repelled by the magnetic field and show a decrease in weight.
A sample of CuS will be attracted by the magnetic field and show an increase in weight.
In the picture below, you can see the sample partially suspended between the poles of an electromagnet.
Answer:
ΔG=ΔG0+RTlnQ where Q is the ratio of concentrations (or activities) of the products divided by the reactants. Under standard conditions Q=1 and ΔG=ΔG0 . Under equilibrium conditions, Q=K and ΔG=0 so ΔG0=−RTlnK . Then calculate the ΔH and ΔS for the reaction and the rest of the procedure is unchanged.
Explanation: