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Anastaziya [24]
3 years ago
11

How many moles are in 2.56 of Hg​

Chemistry
1 answer:
Alexeev081 [22]3 years ago
5 0

Answer: 0.0128 moles

Explanation:

Hg is the chemical symbol for the element, Mercury.

- The molar mass of mercury is 200.6.

- The reacting mass is 2.56g

Recall that the number of moles of any substance = Reacting mass ➗ Molar mass

So, Number of moles = 2.56/200.6

= 0.0128

Thus, there are 0.0128 moles of Mercury (Hg) in 2.56g of mercury

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IrinaVladis [17]

Answer:

the second stage of cell division. its between prophase and anaphase during the chromosomes become attached to spindle fibers.

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3 years ago
How might the yield of 1-bromobutane be affected if water was not added, and what product(s) would be favored?
jek_recluse [69]
  • Due to the inability of the reaction to take place, the yield of 1-Bromobutane would drop.
  • Since 1-Butanol won't react with the additional sodium bromide, bromination won't happen.
  • If water had been supplied, the equilibrium would have shifted extremely far to the left, preventing the reactants from interacting with the acid and favoring the yield of 1-bromobutane instead.
<h3>What is Bromination?</h3>
  • When a substance undergoes bromination, bromine is added to the compound as a result of the chemical reaction.
  • After bromination, the result will have different properties from the initial reactant.
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5 0
2 years ago
A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33
Burka [1]

Answer:

B. A precipitate will form since Q > Ksp for calcium oxalate

Explanation:

Ksp of CaC₂O₄ is:

CaC₂O₄(s) ⇄ Ca²⁺ + C₂O₄²⁻

Where Ksp is defined as the product of concentrations of Ca²⁺ and C₂O₄²⁻ in equilibrium:

Ksp = [Ca²⁺][C₂O₄²⁻] = 2.27x10⁻⁹

In the solution, the concentration of calcium ion is 3.5x10⁻⁴M and concentration of oxalate ion is 2.33x10⁻⁴M.

Replacing in Ksp formula:

[3.5x10⁻⁴M][2.33x10⁻⁴M] = 8.155x10⁻⁸. This value is reaction quotient, Q.

If Q is higher than Ksp, the ions will produce the precipitate CaC₂O₄ until [Ca²⁺][C₂O₄²⁻] = Ksp.

Thus, right answer is:

<em>B. A precipitate will form since Q > Ksp for calcium oxalate</em>

<em></em>

4 0
3 years ago
Which of the following mixtures may be act as a buffer solution?
lions [1.4K]

Answer:

B) HF, NaF

Explanation:

  • A buffer solution is made up of a weak acid and its conjugate base.
  • In this case, HF is a weak acid and NaF is the conjugate base, therefore the mixture of HF,  and NaF would make a perfect buffer solution.
  • The buffer would have the weak acid HF and its conjugate base, F- which comes from the soluble salt NaF.
4 0
3 years ago
What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?
Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>

<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>

<em>∴ The freezing point of the solution is - 4.39 °C.</em>

6 0
3 years ago
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