The term which is described as a long, narrow depression in the ocean floor would be ocean trench. They <span>are hemispheric-scale long but narrow topographic depressions of the sea floor. They are also the deepest parts of the ocean floor. Hope this answers the question.</span>
Answer:
0.5 mole of CO₂.
Explanation:
We'll begin by calculating the number of mole in 42 g of baking soda (NaHCO₃). This can be obtained as follow:
Mass of NaHCO₃ = 42 g
Molar mass of NaHCO₃ = 23 + 1 + 12 + (16×3)
= 23 + 1 + 12 + 48
= 84 g/mol
Mole of NaHCO₃ =?
Mole = mass / molar mass
Mole of NaHCO₃ = 42/84
Mole of NaHCO₃ = 0.5 mole
Next, balanced equation for the reaction. This is given below:
NaHCO₃ + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O + CO₂
From the balanced equation above,
1 mole of NaHCO₃ reacted to produce 1 mole of CO₂
Finally, we shall determine the number of mole of CO₂ produced by the reaction of 42 g (i.e 0.5 mole) of NaHCO₃. This can be obtained as follow:
From the balanced equation above,
1 mole of NaHCO₃ reacted to produce 1 mole of CO₂.
Therefore, 0.5 mole of NaHCO₃ will also react to produce 0.5 mole of CO₂.
Thus, 0.5 mole of CO₂ was obtained from the reaction.
Answer : The correct answer for a) 4-bromo-2-iodo-4-methyl pentane and b)5-bromo-2-ethoxy-2-methyl pentane.
A) Reaction with NaI :
Reaction of alkyl halide with NaI is known as Finkelstein Reaction . The acetone is used as solvent . It involves bimolecular nucleophillic substitution rmechanism (SN²) . There is replecement of one halogen with other occurs .
The incoming Nucleophile(Nu⁻) (halide) attacks on carbon from back side , while the leaving group (halide) leaves the compound from front side , simultaneously. The product so formed have is inverted .(Image)
NaI releases I⁻ ion which act as nucelophile and attacks on C1 carbon and Br⁻ from C1 carbon is released . Out of two bromines at C1 and C4 carbons , C1 is primary carbon which is less sterically hindered while C-4 is tertiary carbon and sterically hindered . So it is easy for incoming Nu⁻ to attack on C1 carbon .So Br⁻ is repleaced by I⁻.
1,4-dibromo-4-methylpentane + NaI → 4-bromo-1-iodo-4-methylpentane
The product formed from reaction between 1,4-dibromo-4-methylpentane and NaI is 4-bromo-1-iodo-4-methylpentane . (Image)
B) Reaction with AgNO3 :
Reaction of alkyl halide with AgNO3 in ethanol takes place via SN¹ ( unimolecular nucleophilic substitution ) mechanism . In this leaving group(halide) leaves from alkyl halide forming an intermediate carbocation species . The incoming Nu⁻ attack on this carbocation.
AgNO3 reacts releases Ag⁺ion which abstract Br⁻ of C-4 carbon from 1,4-dibromo-4-methylpentane. THis forms tertiary carbocation which is more stable than carbocation formed by removal of Br from C-1 . The ethanol being more Nucleophilic than NO₃⁻ (from AgNO₃), attacks on this carbocation .(Image )
The product formed as a result is 5-bromo-2-ethoxy-2-methyl pentane.
The final concentration of the diluted standard is 0.2 mg/dL.
<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>
Using the dilution formula:
where
- C1 is initial concentration
- V1 initial volume
- C2 is final concentration
- V2 is final volume.
Assuming a final volume of 100 mL, and since a 1/5 dilution is made:
C1 = 1.00 mg/dL
V1 = 20
C2 = ?
V2 = 100 mL
C2 = C1V1/V2
C2 = 20 × 1/100
C2 = 0.2 mg/dL
Therefore, the final concentration of the diluted standard is 0.2 mg/dL.
Learn more about dilution at: brainly.com/question/24881505

The emission spectrum of a chemical element or chemical compound is the spectrum of frequencies of electromagnetic radiation emitted due to an atom or molecule making a transition from a high energy state to a lower energy state. The photon energy of the emitted photon is equal to the energy difference between the two states. There are many possible electron transitions for each atom, and each transition has a specific energy difference. This collection of different transitions, leading to different radiated wavelengths, make up an emission spectrum. Each element's emission spectrum is unique. Therefore, spectroscopy can be used to identify elements in matter of unknown composition. Similarly, the emission spectra of molecules can be used in chemical analysis of substances.