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3 years ago

QUESTION 4 Which structure is found in all eukaryotic cells? large central vacuole Golgi apparatus flagella cilia

Chemistry

2 answers:

jok3333 [9.3K]3 years ago

5 0

I believe it's the Golgi Apparatus

mel-nik [20]3 years ago

5 0

Cilia is a <span>structure is found in all eukaryotic cells. </span>

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Suppose 550.mmol of electrons must be transported from one side of an electrochemical cell to another in 49.0 minutes. Calculate

nekit [7.7K]

Answer:

18.0 Ampere is the size of electric current that must flow.

Explanation:

Moles of electron , n = 550 mmol = 0.550 mol

1 mmol = 0.001 mol

Number of electrons = N

$N=N_A\times n$

Charge on N electrons : Q

$Q = N\times 1.602\times 10^{-19} C$

Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds

1 min = 60 seconds

Size of current : I

$I=\frac{Q}{T}=\frac{N\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$=\frac{n\times N_A\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}$

$I=\frac{0.550 mol\times 6.022\times 10^{23} mol^{-1}\times 1.602\times 10^{-19} C}{49.0\times 60 seconds}=18.047 A\approx 18.0 A$

18.0 Ampere is the size of electric current that must flow.

3 0

3 years ago

The most important part of the brain for an organism is the...<br>

Serjik [45]

Answer:

its the lower brainstem

Explanation:

The lowest part of the brainstem, the medulla is the most vital part of the entire brain and contains important control centers for the heart and lungs.

4 0

2 years ago

Part A: You are given a beaker containing a solution of an enzyme that hydrolyzes adenosine triphosphate (ATP) while it generate

Oksi-84 [34.3K]

Answer:

1) magnesium chloride

2) b) The copper is getting oxidized from Cu+ to Cu2+ and turns blue.

Explanation:

The work published by David N. Frick, Sukalyani Banik, and Ryan S. Rypma in J Mol Biol. 2007 Jan 26; 365(4): 1017–1032 clearly shows that divalent metal ions of group 2 such as Mg^2+ play an important role in ATP hydrolysis. Addition of EDTA decreased the rate of hydrolysis of ATP (due to sequestration of the divalent ion of group 2) indicating an active participation of divalent ions in the process.

2) The copper I ion is colourless because it is a d^10 specie. However, when it is oxidized to Cu^2+, a blue colour appears in the solution.

4 0

2 years ago

If you were to react 4.3 miles of magnesium phosphide how many moles of potassium would you need?

bixtya [17]

Answer:

25.8

Explanation:

Let's write the reaction between magnesium-phosphide and potassium:

Mg3P2 + K = Mg + K3P

And now let's balance this equation:

Mg3P2+6K=3Mg+2K3P

We see that the ratio of magnesium-phosphide and potassium is 1:6, which means that for every mole of magnesium-phosphide there need to be 6 moles of potassium.

Since we have 4.3 moles of Mg3P2, there need to be 6 • 4.3 = 25.8 moles of potassium.

5 0

2 years ago

What would be the major product if 1,4-dibromo-4-methylpentane was allowed to react with:

Levart [38]

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.

7 0

3 years ago