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Spectroscopy be used to distinguish between the following is the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
<h3>What is spectroscopy?</h3>
Spectroscopy is the study of emission or absorption of light. It is used to study the structure of atoms and molecules.
The three types of spectroscopy are:
- atomic absorption spectroscopy (AAS)
- atomic emission spectroscopy (AES)
- atomic fluorescence spectroscopy (AFS)
Thus, the correct option is B, the compound B has a peak at 3200 – 3500 cm⁻¹ in its IR spectrum.
Learn more about spectroscopy
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Answer:
answer
Explanation:
sorry I do not know the answer, have a great day
Answer:
Because both CaCl2 and CaBr2 both contain elements (Chlorine and Bromine) from the same group (group 7)
Explanation:
Elements are placed into different groups in the periodic table. Elements in the same group are those that have the same number of valence electrons in their outermost shell and as a result will behave similar chemically i.e. will react with other elements in the same manner.
Chlorine and Bromine are two elements belonging to group 7 of the periodic table. They are called HALOGENS and they have seven valence electrons in their outermost shell. Hence, when they form a compound with Calcium, a group two element, these compounds (CaCl2 and CaBr2) will possess similar properties because they have elements that are from the same group (halogen group).
<h3>
Answer:</h3>
0.127 mol Au
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 25.0 g Au
[Solve] moles Au
<u>Step 2: Identify Conversions</u>
[PT] Molar Mass of Au - 196.97 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Multiply/Divide [Cancel out units:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
0.126923 mol Au ≈ 0.127 mol Au
