Answer:
412 g Cl₂
General Formulas and Concepts:
<u>Atomic Structure</u>
- Reading a Periodic Table
- Moles
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
[Given] 3.50 × 10²⁴ molecules Cl₂
[Solve] grams Cl₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Cl - 35.45 g/mol
Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Divide/Multiply [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
412.072 g Cl₂ ≈ 412 g Cl₂
Answer:
1-ethyl-2-methyl cyclopropane.
Explanation:
- The structure of the molecule will be as shown in the attached image.
- The molecular formula of the compound is C₆H₁₂.
- It has 3 membered ring with 3 C atoms and two substituents one of them with one C atom (methyl) and the other with 2 C atoms (ethyl).
- The ring consist of 3 C atoms, so its name is cyclo propane.
- We numbering the atoms of the ring that give the ethyl substituent the low no. (1) and then methyl group take no. (2).
- <em>Thus, the name of the compound is 1-ethyl-2-methyl cyclopropane.</em>
Answer:
Hello There!!
Explanation:
I believe the answer is nuclear.
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The concentration of HCl is equal to 2.54mol/L.
<h3>Mole calculation</h3>
To solve this question, one must use the molarity calculation, which corresponds to the following expression:

Thus, to find the molarity of the sample, the following calculations must be performed:



So, 0.00254 moles were added per 10ml, so we can do:

So, the concentration of HCl is equal to 2.54mol/L.
Learn more about mole calculation in: brainly.com/question/2845237
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration
.


The given equilibrium reaction is,

Initially c 0
At equilibrium

The expression of
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)

where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:



Therefore, the value of equilibrium constant for this reaction is, 1.1