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2 years ago

How the proton H gradient is used to make ATP

Chemistry

1 answer:

sladkih [1.3K]2 years ago

4 0

Answer:

The proton gradient produced by proton pumping during the electron transport chain is used to synthesize ATP. Protons flow down their concentration gradient into the matrix through the membrane protein ATP synthase, causing it to spin (like a water wheel) and catalyze conversion of ADP to ATP.

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A compound composed of 3. 3 % h, 19. 3 % c, and 77. 4 % o has a molar mass of approximately 60 g/mol. What is the molecular form

Brut [27]

The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as Carbonic acid.

<h3>What is empirical formula and molecular formula?</h3>

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

O = 77.4%

No. of moles of H = 3.3/1

No. of moles of H = 3.3

No. of moles of C = 19.3 / 12

No. of moles of C = 1.60

No. of moles of O = 77.4/16

No. of moles of O = 4.83

Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60

The ratio of the atoms of C, H and O = 2 : 1 : 3

So, the empirical formula is $\mathrm{H}_{2} \mathrm{CO}_{2}$

Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

Then, molar mass $$=(2 \times 1+1 \times 12+3 \times 16) n\\$

Molar mass = 62n

As the question, 62 n = 60

n = 0.96 or n = 1 (rounded off to nearest ones)

So, the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$ i.e., the compound is Carbonic acid.

To know more about molecular formula visit:

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7 0

1 year ago

Which is chemical property of vinegar

kobusy [5.1K]

Acidic? Vinegar is also known as ethanoic acid, being a carboxylic acid. It has a low pH value compared to alkaline substances (:

7 0

3 years ago

Chromium(III) oxide reacts with hydrogen sulfide (H2S) gas to form chromium(III) sulfide and water: Cr2O3(s) + 3H2S(g) → Cr2S3(s

Tpy6a [65]

Answer:

(a) 0.779 mol Cr₂O₃

(b) 118 g Cr₂O₃

Explanation:

Step 1: Convert 156 g Cr₂S₃ to moles

Cr Molar Mass - 52.00 g/mol × 2 = 104.00 g/mol

S Molar Mass - 32.07 g/mol × 3 = 96.21 g/mol

156 g Cr₂S₃ ÷ 200.21 g/mol = 0.779182 mol Cr₂S₃

Step 2: Find conversion from Cr₂S₃ to Cr₂O₃

1 mol Cr₂S₃ equals 1 mol Cr₂O₃

Step 3: Use Dimensional Analysis

0.779182 mol Cr₂S₃ · $\frac{1 \hspace{2} mol \hspace{2} Cr_2O_3}{1 \hspace{2} mol \hspace{2} Cr_2S_3}$ = 0.779182 mol Cr₂O₃

0.779182 mol Cr₂O₃ ≈ 0.779 mol Cr₂O₃ (3 significant figures)

Step 4: Convert moles to grams

Cr Molar Mass - 52.00 g/mol × 2 = 104.00 g/mol

O Molar Mass - 16.00 g/mol × 3 = 48.00 g/mol

0.779182 mol Cr₂O₃ · 152.00 g/mol = 118.436 g Cr₂O₃

118.436 g Cr₂O₃ ≈ 118 g Cr₂O₃ (3 significant figures)

6 0

2 years ago

30g of naoh is dissolved in 1.5 liter solution the active mass of naoh is?

Pepsi [2]

Answer:

0.5mol/L

Explanation:

First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the question = 30g

Number of mole = Mass /Molar Mass

Number of mole = 30/40 = 0.75mol

Volume = 1.5L

Active mass = mole/Volume

Active mass = 0.75mol/1.5L

Active mass = 0.5mol/L

7 0

3 years ago

How many particles are present in 12.47 grams of NaCl

liq [111]

Answer:

1.26*10²³ particles are present in 12.47 grams of NaCl

Explanation:

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. The Avogadro number applies to any substance.

So, first of all you must know the amount of moles that represent 12.47 grams of NaCl. For that it is necessary to know the molar mass.

You know:

Na: 23 g/mole
Cl: 35.45 g/mole

So the molar mass of NaCl is: 23 g/mole + 35.45 g/mole= 58.45 g/mole

Now you apply a rule of three as follows: if 58.45 grams are present in 1 mole of NaCl, 12.47 grams in how many moles will they be?

$moles=\frac{12.47 grams*1 mole}{58.45 grams}$

moles= 0.21

You apply a rule of three again, knowing Avogadro's number: if in 1 mole of NaCl there are 6,023 * 10²³ particles, in 0.21 moles how many particles are there?

$number of particles=\frac{0.21 moles*6.023*10^{23} }{1 mole}$

number of particles= 1.26*10²³

<u><em>1.26*10²³ particles are present in 12.47 grams of NaCl</em></u>

5 0

3 years ago