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PilotLPTM [1.2K]
3 years ago
10

Helium can be liquefied when he atoms are attracted to one another by intermolecular ________ forces.

Chemistry
1 answer:
Akimi4 [234]3 years ago
3 0

Helium can be liquefy through a very low temperature because of the weakness of attractions between the helium atoms. In addition, helium is a noble gas that has a very weak interatomic London dispersion forces. Thus, this element would remain liquid at atmospheric pressure all the way to its liquefaction point going to absolute zero.

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What element is being oxidized in the following redox reaction?
GrogVix [38]

Answer:

The O is being oxidized, but at the same time, is being reducted.

Explanation:

H₂O₂(l) + ClO₂(aq) → ClO₂(aq) + O₂(g)

In this reaction, we have 4 compounds:

Hydrogen peroxide

Chlorine dioxide (twice)

Oxygen

In both dioxide, the Cl acts with +4 in oxidation state; the oxygen acts with -2.

Oxgen in ground state has 0, as oxidation number.

In peroxide, the H acts with +1 but  the oxygen acts with -1.

Peroxide is making the oxidation number from the O in the ClO₂, to decrease (reduction) and to increase in the O, at the ground state.

Hydrogen peroxide is a good reducing and oxidizing agent at the same time.

4 0
3 years ago
Toolkit <br> How many energy levels does sulfur have?<br> A. 2<br> B. 3<br> C. 4<br> D. 14
True [87]
Sulfur has 3 energy levels. 
5 0
3 years ago
Calculate the theoretical yield for the bromination of both stilbenes
podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

4 0
2 years ago
Which statement about the reaction is correct?
Goshia [24]

The reactions based on the absorption and release of the energy are called endothermic and exothermic reactions. The reaction is exothermic.

<h3>What is an exothermic reaction?</h3>

Exothermic reactions are the reaction in which the reactant produces products that release energy from the system to the surroundings. In the reaction bond energy of the reactant is less than the product.

Energy from the system is released in the form of heat, sound, light and electricity. The weak bonds of the compounds are replaced with stronger ones and the standard enthalpy of the reaction is negative.

Therefore, option c. reaction is exothermic is correct.

Learn more about the exothermic reactions here:

brainly.com/question/26616927

7 0
2 years ago
Read 2 more answers
Addition of water to an alkyne gives a keto‑enol tautomer product. Draw an enol that is in equilibrium with the given ketone
Marrrta [24]

Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.

<h3>What is the keto-enol means in tautomer?</h3>

They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.

  1. The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
  2. If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
  3. If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
  4. This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.

7 0
2 years ago
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