Plsss help with this ASAP TT

Answer:

FADH2 has a lower (less negative) redox potential than NADH does

Explanation:

Flavin Adenine Dinucleotide (FAD) and Nicotinamide Adenine Dinucleotide (NAD) are redox cofactors that play important functions for mitochondrial activity and cellular redox balance. Both coenzymes exist in two forms: an oxidized and a reduced, which are abbreviated as NAD/FAD and NADH/FADH2, respectively. These reduced forms (NADH and FADH2) are produced in the Krebs cycle during respiration. FADH2 has lower redox potential than NADH because FADH2 is only capable of activating 2 proton pumps, while NADH can activate 3 proton pumps during the electron transport chain, thereby FADH2 generates a minor number of ATP molecules than NADH.

[tex]1.44\times 10^{-2}[/tex]

Explanation:

We need to find the value of [tex]3.49\times 10^{-3}+1.1\times 10^{-2}[/tex].

Here,

The first number is [tex]3.49\times 10^{-3}[/tex]

The second number is [tex]1.1\times 10^{-2}[/tex]

To add these numbers as follows :

[tex]3.49\times 10^{-3}+1.1\times 10^{-2}=0.01449\\\\=1.44\times 10^{-2}[/tex]

So, the value of [tex]3.49\times 10^{-3}+1.1\times 10^{-2}[/tex] is [tex]1.44\times 10^{-2}[/tex]

Answer:

The third one

Explanation:

An ore is a naturally occurring solid material from which a metal or valuable mineral can be profitably extracted.

Answer:

35.453 g

Explanation:

one mole of chlorine is 34.453 g

Answer:

One mole of chlorine atoms has a mass of 35.453g. A mole of something means that you have 6.023×1023 units of that something. So if you have a mole of Cl atoms you have 6.023×10  

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Answer:

1.65485688 × 10121 grams

Explanation:

Aside from those discrete items, anything like rice or flour can be measured out to make up 100 grams. For example, 1/2 cup (125 ml or about 4 and 1/2 fl. oz) of dry rice weighs very close to 100 grams.

Answer:

2.76 gram

Explanation:

2AI + 6HCl → 2AlCl3 + 3H2

Answer

Al +2 HCl - AlCl2 + H2

Explanation:

eam=%abudance×mass+%abudance×mass

eam=19.8%×10.013/100+80.2%×11.009/100

eam=198.2574/100+882.9218/100

eam=1081.1792/100

eam=10.811792

eam=10.812

Answer:

0.1828g/cm³

Explanation:

density= mass÷volume

m= 12.6g

v= 4.1×4.1×4.1 = 68.921cm³

•

density= 12.6÷68.921 = 0.1828g/cm³

Answer:

[tex]\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}[/tex]

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:                      98.08           392.18

             2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

[tex]\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}[/tex]

(ii) Moles of H₂SO₄

[tex]\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}[/tex]

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ [tex]\text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}[/tex]

(iv) Mass of Cr₂(SO₄)₃ [tex]\text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}[/tex]

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

[tex]\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}[/tex]

Answer:

The answer is "2.78".

Explanation:

Given values:

CD= 0.44

Formula:

[tex]\bold{f_r= \frac{v^{2}}{ g(s-1)D}}[/tex]

g=9.8

s= 2.7

D= 0.06

[tex]\to f_r=\frac{2.5^2}{9.8(2.7-1)0.06}\\\\[/tex]

        [tex]=\frac{2.5 \times 2.5}{9.8 \times 1.7 \times 0.06 }\\\\=\frac{6.25}{.9996 }\\\\=6.252501[/tex]

[tex]\phi = \frac{82\times v}{ \sqrt{cD} \times f_r^{-1.5}}\\\\[/tex]

   [tex]= \frac{82 \times 0.25 }{ \sqrt{0.44} \times 6.25^{1.5}}\\\\=2.4285\\[/tex]

[tex]\frac{\bigtriangleup P f_1 s_1}{L} = \frac{\bigtriangleup Pf_w}{L}(1+\phi)\\[/tex]

           [tex]=\frac{2fwSwv^2 (1+2.4285)}{D}\\\\[/tex]

[tex]Re= \frac{D \bar v Sw}{M_w}\\[/tex]

     [tex]=\frac{0.06 \times 2.5 \times 1000 }{0.001}\\\\=\frac{150 }{0.001}\\\\= 150 \times 10^{3}\\\\= 1.50 \times 10^{5}\\\\[/tex]

[tex]fw= 0.00389[/tex]

[tex]\to \frac{\bigtriangleup P f_1 s_1}{L}[/tex]

[tex]\to \frac{2 \times 0.00389 \times 1000 \times2.5^2 \times 3.4265}{0.06}\\\\\to 2.78[/tex]

[tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex] is the correct electron configuration for an element with five electrons in the third energy level.

Elements are the simplest substances which cannot be broken down using chemical methods.

The shell nearest to the nucleus, 1n, can carry two electrons, while the next shell, 2n, can carry eight, and the third shell, 3n, can carry up to eighteen.

The third shell carries 18 electrons; 2 in a 3s orbital; 6 in three 3p orbitals; and 10 in five 3d orbitals. The fourth shell carries  32 electrons; 2 in a 4s orbital; 6 in three 4p orbitals; 10 in five 4d orbitals; and 14 in seven 4f orbitals.

The element would be phosphorus. Its electron configuration is [tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex]

Learn more about elements here:

brainly.com/question/1896898

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