If there is an increase in industrial activity, that means that more heat will be dissipated to the atmosphere in the form of carbon dioxide. Industrialization requires fuel to keep the processes on the go. At the end of the pipeline, the combustion of fuel would result to carbon dioxide released to the atmosphere. That's how it is contributing to the global climate change through the greenhouse effect.
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S and S²⁻ do not have the outer subshell fully filled with electrons.
Explanation:
We look at electronic configurations:
Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons
S 1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons
Zn²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons
S²⁻ 1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons
Ca²⁺ 1s² 2s² 2p⁶ 3s² 3p⁶ - the outer subshell 3p⁶ is fully-filled with electrons
Learn more about:
electron configurations
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Answer:
![\boxed{\mathrm{Alkene}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cmathrm%7BAlkene%7D%7D)
Explanation:
The hydrocarbon shown has a double bond. Hydrocarbons with double bonds are known as alkenes.
Cyclic alkanes have cyclic structure.
Alkanes only have single bonds.
Alkynes have triple bonds.
Answer:
6.82 g H₂S
General Formulas and Concepts:
<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>
<u>Stoichiometry</u>
- Using Dimensional Analysis
Explanation:
<u>Step 1: Define</u>
0.200 mol H₂S
<u>Step 2: Identify Conversions</u>
Molar Mass of H - 1.01 g/mol
Molar Mass of S - 32.07 g/mol
Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol
<u>Step 3: Convert</u>
- Set up:
![\displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.200%20%5C%20mol%20%5C%20H_2S%28%5Cfrac%7B34.09%20%5C%20g%20%5C%20H_2S%7D%7B1%20%5C%20mol%20%5C%20H_2S%7D%29)
- Multiply:
![\displaystyle 6.818 \ g \ H_2S](https://tex.z-dn.net/?f=%5Cdisplaystyle%206.818%20%5C%20g%20%5C%20H_2S)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
6.818 g H₂S ≈ 6.82 g H₂S