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3 years ago

What is the relationship between Ka and Kb with Kw?

2 answers:

lina2011 [118]3 years ago

3 0

The correct answer is ...... The product of Ka and Kb equals the auto-dissociation constant for water. I know its right for sure!

Eduardwww [97]3 years ago

3 0

Answer is: The product of Ka and Kb equals the auto-dissociation constant for water.

Reaction of auto-dissociation for water: H₂O ⇄ H⁺ + OH⁻.

Water dissociates (autoionization) to form hydrogen ions (H⁺) and hydroxide (OH⁻) ions. The protons (H⁺) hydrate as hydroxonium ions( H₃O⁺).

The Kw (the ionic product for water) at 25°C is 1·10⁻¹⁴ mol²/dm⁶ or 1·10⁻¹⁴ M². Concentration of hydrogen ions and hydroxide ions in pure water are the same.

Ka · Kb = Kw; the ionic product of water.

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Two different atoms have four protons each and the same mass. However, one has a positive charge while the other is neutral. Des

seraphim [82]

The possible number and location of all subatomic are one of them is electrically neutral, while the other has a stable electronic configuration.

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8 0

2 years ago

Which of the following questions is a scientific question?

insens350 [35]

Explanation:

where's the question, please?

4 0

3 years ago

Adenosine triphosphate (ATP) is the main energy currency used in cells. ATP hydrolysis is coupled with unfavorable reactions, ma

kotegsom [21]

Answer:

Thioesters- the Sulfur-Carbon bond is hydrolyzed, e.g. AcetylCoA

Reduced cofactor- These compounds accept electrons during the oxidation of substrates and energy is released when they are oxidized, e.g. Ubiquinol

Phosphorylated compounds- These compounds yield H₂P0₄⁻ upon hydrolysis, e.g. Phosphocreatine

Explanation:

Thioesters are esters in which the linking oxygen atom is replaced by a sulphur atom. They are the product of esterification between a carboxylic acid and a sulfhydryl group (thiol). They have the functional group R–S–CO–R'. Thioesters are common intermediates in many biosynthetic reactions. Examples include malonyl-CoA, acetyl-CoA, and propionyl-CoA. In the hydrolysis of thioesters, the sulfur-carbon bond is hydrolyzed.

Cofactors can be either organic or inorganic molecules that are required by enzymes to function. Cofactors can be oxidized or reduced for the enzymes to catalyze the reactions. Examples include, NAD, FAD, NADP, Coenzyme Q₁₀. Coenzyme Q₁₀ exists in three redox states, fully oxidized, partially reduced, and fully reduced. Ubiquinol is the reduced (electron-rich) form of coenzyme Q₁₀. Coenzyme Q₁₀ is vital for proper transfer of electrons within the mitochondrial oxidative respiratory chain, whose main function is to produce adenosine triphosphate (ATP). These compounds accept electrons during the oxidation of substrates and energy is released when they are oxidized.

Phosphorylated compounds are compounds with a phosphoryl group (PO₃²⁻) attached to its molecules. These compounds yield an inorganic phosphate (H₂P0₄⁻) upon hydrolysis. Phosphorylation is especially important for protein function as this modification activates or deactivates almost half of the enzymes, thereby regulating their function. Examples include, glucose-1-phosphate, phosphoserine, phosphocreatine, etc.

4 0

3 years ago

How many electrons will calcium gain or lose when it forms an ion?

Alexeev081 [22]

Calcium will lose 2 electrons to gain a full outer shell.
This will form a positive ion.

4 0

3 years ago

Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4

kozerog [31]

Answer:

The feed ratio (liters 20% solution/liter 60% solution) is 3,08

Explanation:

In this problem you have a 20,0 wt% H₂SO₄ and a 60,0 wt% H₂SO₄ solutions.

100 kg of 20% solution are 100kg/1,139 kg/L = 87,8 L 

100kg×20wt% = 20 kg H₂SO₄. In moles:

20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ 203,9 mol

The final molarity 4,00M comes from:

$\frac{203,9 moles+ Xmoles}{87,8L + Yliters}$ (1)

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L× $\frac{1,498 kgsolution}{L}$ × $\frac{60 kg H_{2}SO_{4}}{100kgSolution}$ × $\frac{1kmol}{98,08kg H_{2}SO_{4}}$ = 0,9164 kmolH₂SO₄ ≡ 916,4 moles

That means:

X/Y = 916,4/100 = 9,164 (2)

Replacing (2) in (1):

Y(liters of 60,0 wt% H₂SO₄) = 28,52 L

Thus, feed ratio (liters 20% solution/liter 60% solution):

87,8L/28,52L = 3,08

I hope it helps!

8 0

3 years ago