Answer:
Explanation:
Hello,
In this case, since the Lewis theory is based on the bonds formation between atoms via the valence electrons, we can verify the chemical formula of the compound formed by strontium and nitrogen by noticing that strontium has two valence electrons as it is in group IIA, for that reason, two nitrogens should be available for bonding. Therefore, since nitrogen is in group VA, it is said that three electrons are required to attain the octet (maximum amount bonded electrons), for that reason, three strontiums are should be available for bonding. In such a way, the formula should be:
Regards.
The question is incomplete, here is the complete question:
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹
Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
<u>Answer:</u> The concentration of 
in the vessel after 0.240 seconds is 0.24 M
<u>Explanation:</u>
For the given chemical equation:
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken= 0.240 second
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 1.44 M
Putting values in above equation, we get:
![14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)](https://tex.z-dn.net/?f=14.1%3D%5Cfrac%7B1%7D%7B0.240%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B1.44%7D%5Cright%29)
![[A]=0.245M](https://tex.z-dn.net/?f=%5BA%5D%3D0.245M)
Hence, the concentration of in the vessel after 0.240 seconds is 0.24 M
Answer:
3–heptene or Hept–3–ene.
Explanation:
To name the compound given in the question above, we must determine the following:
1. Identify the functional group of the compound.
2. Locate the longest continuous carbon chain. This gives the parent name of the compound.
3. Identify the substituent group attached to the compound.
4. Locate the position of the double bond (i.e the functional group) by giving it the lowest possible count.
5. Combine the above to obtain the name of the compound.
Now, we shall obtain the name of the compound as follow:
1. The compound contains double bond. Hence the compound is an alkene.
2. The longest continuous carbon chain is 7 i.e heptene.
3. No substituent group is attached to the compound.
4. The double bond is located at carbon 3 since counting from the right gives the lowest count for the double bond.
5. The name of the compound is:
3–heptene or Hept–3–ene
<h3>
Answer:</h3>
296 g BaBr₂
<h3>
General Formulas and Concepts:<u>
</u></h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<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
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 8.56 × 10²³ formula units BaBr₂
[Solve] grams BaBr₂
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Ba - 137.33 g/mol
[PT] Molar Mass of Br - 35.45 g/mol
Molar Mass of BaBr₂ - 137.33 + 2(35.45) = 208.23 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>
295.99 g BaBr₂ ≈ 296 g BaBr₂
Thank you agenthammerx for helping me with this question!
Answer:
A digital camera. You can zoom and all that stuff and see individual pixels when close up. Also none of the other options would make sense.
Explanation:
