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olchik [2.2K]
2 years ago
15

As the electron transport carriers shuttle electrons, they actively pump _____ into the outer membrane compartment, setting up a

concentration gradient called the proton motive force.
Chemistry
1 answer:
balu736 [363]2 years ago
8 0

Answer:

Hydrogen ions or protons

Explanation:

Electron transport carriers is a series of complexes that transfer electrons from electron donors to electron acceptors via redox reactions, and couples this electron transfer with the transfer of protons (H+ ions) across a membrane. This creates an electrochemical proton gradient that drives the synthesis of ATP, a molecule that stores energy chemically in the form of highly strained bonds. The molecules of the chain include peptides, enzymes (which are proteins or protein complexes), and others. The final acceptor of electrons in the electron transport chain during aerobic respiration is molecular oxygen although a variety of acceptors other than oxygen such as sulfate exist in anaerobic respiration.

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A 600.0 mL sample of 0.20 MHF is titrated with 0.10 MNaOH. Determine the pH of the solution after the addition of 600.0 mL of Na
Leona [35]

Answer: pH=12.69

Explanation:

{\text{Moles of HF}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of HF}=0.20M\times 0.6L=0.12 moles

HF\rightarrow H^++F^-

Initial 0.12               0       0

Eqm   0.12-x           x        x

K_a=\frac{[H^+][F^-]}{[HF]}

3.5\times 10^{-4}=\frac{x^2}{0.12-x}  

(neglecting small value of x in comparison to 0.12)

x=4.2\times 10^{-5}

Moles of H^+=4.2\times 10^{-5}

NaOH\rightarrow Na^++OH^-

{\text{Molesof NaOH}}=Molarity\times {\text{Volume of solution in liters}}

{\text{Moles of NaOH}}=0.10M\times 0.6L=0.06 moles

0.06 moles of NaOH will give 0.06 moles of [OH^-]

Now 4.2\times 10^{-5} moles of OH^- will be neutralized by 4.2\times 10^{-5} moles of H^+ and (0.06-4.2\times 10^{-5})=0.059 moles of OH^- will be left.

Molarity of OH^-=\frac{0.059moles}{1.2L}=0.049M

pOH=-\log[OH^-]=-\log[0.049]=1.31

pH = 14 - pOH= 14 - 1.31 = 12.69

5 0
2 years ago
The mass unit associated with density is usually grams. if the volume (in ml or cm3) is multiplied by the density (g/ml or g/cm3
jok3333 [9.3K]

The weight in grams = 7.93 g

Given volume = 2.00 in^{3}

Given  density = 0.242 g/cm^{3}

We need to find the Mass(weight) in grams.

To find the weight in grams we need to keep in mind that the volume and density must use the same volume unit for cancellation. So that the volume units will cancel out, leaving only the mass units.

The unit of given volume is in^{3} and unit of volume in density is cm^{3} , so first we need to change the unit of volume from in^{3} to cm^{3} so that the volume units will cancel out, leaving only the mass units.

1 in^{3} = 16.39 cm^{3} (given conversion)

2 in^{3}\times \frac{(16.39 cm^{3})}{(1 in^{3})}

in^{3} units get cancel out leaving the cm^{3} unit.

2 in^{3} = 32.78 cm^{3}

Mass = Density X Volume.

Density = 0.242 g/cm^{3} and Volume = 32.78 cm^{3}

Mass =0.242\frac{g}{cm^{3}}\times 32.78 cm^{3}

Mass = 7.93 grams (g)


3 0
1 year ago
Why is mining for coal a long-term concern?
Free_Kalibri [48]
A. Fossil fuels take a very long time to form so we mine it faster than it can replenish.
8 0
1 year ago
Read 2 more answers
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
2 years ago
What volume (ml) of a 0.2450 m koh(aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4(aq) solution?
Nonamiya [84]
<span>Answer: It depends on what came after "0.5440 M H...". If it was a monoprotic acid, like HCl, the calculation would go like this: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH If it was a diprotic acid, like H2SO4, like this: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH If it was a triprotic acid, like H3PO4, like this: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH</span>
5 0
2 years ago
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