D - Silicon
This is because Silicon has four electrons as indicated by the group it's in. Chlorine would have 7, Magnesium would have 2, Phosphorous would have 5, and Sulfur would have 6.
Hope this helps!
Chlorophyll, it is the pigment in plants that absorbs light in photosynthesis
Cyclohex-2-en-1-one product would be obtain from intramolecular aldol condensation of 5-oxohexanal.
<h3>What is Intramolecular Aldol Condensation ? </h3>
The condensation reaction in which two ketone groups and aldehyde group in the same molecule are called Intramolecular Aldol Condensation. Intramolecular Aldol Condensation occurs in five or six membered α, β- unsaturated aldehyde or ketones are formed.
Thus from the above conclusion we can say that Cyclohex-2-en-1-one product would be obtain from intramolecular aldol condensation of 5-oxohexanal.
Learn more about the Aldol Condensation here: brainly.com/question/14260386
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Answer:
3.329 g
Explanation:
First you need to determine the molar mass of H2S which is 34.1 g/mol.
With that we know that to find the moles of H2S we just divide the mass of sample with the molar mass.
3.54 g / 34.1 g/mol = 0.103812317 mol of H2S
This means that there is also 0.103812317 mol of sulfur since there is 1 mole of sulfur per 1 mole of H2S.
The molar mass of sulfur is 32.065 g/mol and to find the mass of sulfur you need to multiply the molar mass with the moles of the compound.
0.103812317 mol * 32.065 g/mol = 3.329 g of sulfur
Let me know if you get something else or if something is unclear in the comments so that we can figure it out.
Answer:
See Explanation
Explanation:
a. pH of 1M HOAc(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 1.0M x x
Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵
=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M
=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37
b. pH of 0.10M CH₃NH₃OH(aq)
CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴
C(eq) 0.10M x x
=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M
=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M
=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18
=> pH = 14 – pOH = 14 – 2.18 = 11.82
c. pH of 0.30M HOAc/0.10M OAcˉ(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 0.30M x 0.10M
=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]
= 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M
=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26