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2 years ago

Oxidative phosphorylation could not occur without glycolysis and the citric acid cycle, because _____.

Chemistry

1 answer:

katrin [286]2 years ago

8 0

Answer:

these two processes supply the electrons that are needed for the electron transport chain

Explanation:

Oxidative phosphorylation is the process in which electrons transfer from electron donors to electron acceptors (usually oxygen). These reactions are called redox reactions, and they provide energy used to form ATP.

Electron donors (NADH and FADH2) used in oxidative phosphorylation are produces in some of the catabolic biochemical processes, such as glycolysis, the citric acid cycle, and beta oxidation. The NADH and FADH2 are energy-rich molecules because each of them contains a pair of electrons thus having a high transfer potential. Because of that, oxidative phosphorylation could not happen without first obtaining electron donors in glycolysis and citric acid cycle.

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how many calories of energy are given off to lower the temperature of 100.0g of iron from 150.0°C to 35.0°C?

jeyben [28]

Answer:

ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C

given the specific heat of iron as 0.108 cal/g·°C

heat=(100.0 g)(0.108 cal /g· °C )(125°C) =

100x 0.108x125= 1350 cal

4 0

3 years ago

Changing the number of protons in an atom makes<br><br> A. an ion<br> B. an isotope

lana [24]

A. an ion

The atom gains a net electrical charge if the number of protons and electrons are not equal which makes it an ion.

5 0

2 years ago

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

Best regards.

5 0

3 years ago

Read 2 more answers

Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.

Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= $moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.241}{0.0137kg}=17.6mole/kg$

Therefore, the molality of solution is 17.6 mole/kg

3 0

3 years ago

Scientist can use trees to look at climates of the past. How? What information can they gather and how do they gather it? Explai

andrezito [222]

Explanation:

Scientist use trees a whole lot to look at climate of the past by examining tree rings.

These are layers of cambium in each successive years formed. They have an annual growth pattern and are known as tree rings.

Tree rings can be used to decipher the age of a tree.

These three rings can be used to interpret climatic patterns.
During a wet climate, the tree rings are more robust and bigger.
In a dry climate, the rings are thinner.
These alternating patterns can be used to decipher the climatic signatures in a tree.
Sometimes, it is possible to evaluate some certain isotopes that are useful in climatic studies.

learn more:

Climate change brainly.com/question/7824762

#learnwithBrainly

6 0

2 years ago