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dimaraw [331]
3 years ago
9

Which of the following is true of atoms with very low electronegativity

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

Answer:

well there is no chose here is my best option i would chose

1 Electronegativity is the ability of an atom to attract electrons.

Explanation:

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What is the product of the reaction of hydrobromic acid and 2-bromo-1-butene in the presence of acid and ether?
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Answer: The major product of the reaction between Hydrobromic Acid and 2-bromo-1-butene in the presence of ether and acid is 2,2-dibromobutane.

Explanation:

The mechanism of the reaction is supported by the Markovnikov's rule which explains that in the addition reaction of alkenes by hydrogen-halogen compounds, the incoming halogen substituent goes to the more substituted Carbon. It can also be stated that incoming hydrogen atom goes to the Carbon with more Hydrogen atoms.

The only case when the reverse of Markovnikov's rule takes place is when Hydrogen peroxide is present and the addition reagent is HBr.

This case is not like that and it simply follows the Markovnikov's rule.

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What is the 3-dimensional orientation of a sublevel known as?
RoseWind [281]

The 3-dimensional orientation of a sublevel is known as atomic orbital.

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Each sublevel has an orbital and it is oriented differently in 3-dimensional space.

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2 years ago
Which are balanced and which are unbalanced?
erica [24]

A. is balanced

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3 0
3 years ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
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