The oxidation state of the elements in the compounds are:
CoH₂:
FeBr₃:
<h3>What is the oxidation states of the elements in the given compounds?</h3>
The oxidation states of the elements in each of the given compounds is determined as follows:
Cobalt dihydride, CoH₂
Co = +2
H = -1
Iron (iii) bromide, FeBr₃
Fe = +3
Br = -1
In conclusion, the oxidation state of the elements are charges they have in the compound.
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Answer:
The question is incomplete as some details are missing. Here is the complete question ; A chemist adds 45.0mL of a 0.434M copper(II) sulfate CuSO4 solution to a reaction flask. Calculate the mass in grams of copper(II) sulfate the chemist has added to the flask. Round your answer to 2 significant digits
Explanation:
The step by step explanation is as shown in the attachment
Fats are large molecules made of two types of molecules, glycerol and some type of fatty acid.
The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
<h3>
What is Mole ?</h3>
A mole is a very important unit of measurement that chemists use.
A mole of something means you have 6.023 x 10 ²³ of that thing.
- For 2.15 mol of hydrogen sulphide (H₂S) :
1 mole hydrogen sulphide (H₂S) = 34.08088 grams
Therefore,
2.15 mol of hydrogen sulphide (H₂S) = 34.08088 grams x 2.15 mol
= 73.272 gm
- For 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) ;
1 mol of lead(II) iodide, (PbI₂) = 461.00894 grams
Therefore,
3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) = 461.00894 grams x 3.95 × 10⁻³ mol
= 1.82 gm
Hence,The mass of 2.15 mol of hydrogen sulphide (H₂S) will be 73.272 gm and the mass of 3.95 × 10⁻³ mol of lead(II) iodide, (PbI₂) will be 1.82 gm
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Um to my best knowledge I believe it is the carbonated water
