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vfiekz [6]
3 years ago
7

1 mol Fe2(CrO4)3=____g

Chemistry
1 answer:
IgorLugansk [536]3 years ago
7 0
111.6g of Fe + 156g of Cr + 192g of O = 459.6g of Fe2(CrO4)3
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Ions in solution are found in a hydration shell of water. This shell must be removed for an ion to pass through the channel. How
kirill [66]

The K+ channel uses the energy in ATP hydrolysis to remove the hydration shell from the K+ion

All of this truly implies is that one of these bonds breaks in a hydrolysis process (a water-mediated breakdown reaction), releasing a sizable amount of energy.

The following reaction results in the hydrolysis of ATP to ADP:

ATP+H2O⇋ADP+Pi+energy

The hydrolysis of ATP to ADP is reversible, like the majority of chemical processes.

Energy is needed for the reverse process, which creates ATP from ADP and Pitext P iPi start subscript, P, end subscript, start subscript, I end subscript.

Because cells frequently use (hydrolyze) ATP molecules and depend on constant production of replacement ATP, ATP regeneration is crucial1^1.

Hence the  K+ channel uses the energy in ATP hydrolysis to remove the hydration shell from the K+ion

Learn more about hydrolysis here:

brainly.com/question/4352413

#SPJ4

6 0
2 years ago
Calculate the mass of NaF that must be added to 300ml if a 0.25m HF solution to form a buffer solution with a ph of 3.5​
Kamila [148]

Answer:

6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5

Explanation:

For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).

Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.

HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴

Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).

For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)

= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.

Check using the Henderson - Hasselbalch Equation...

pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18

Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.

pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.

One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.

4 0
3 years ago
Select the characteristics of a hydrogen bond:
Volgvan
I think the answer would be A
6 0
3 years ago
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Chemical name of F3S2
natima [27]

Answers;

F3S2 Name: Iron Disulfide

Lead(II) nitrate formula: Pb(NO3)2

Silicon tetrafluoride formula: SiF4

4 0
2 years ago
Which of the following is not an example of a molecule?
satela [25.4K]

Mn is the answer i just took the test

6 0
3 years ago
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