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Darya [45]
3 years ago
5

What type of bond is present in NBr?

Chemistry
1 answer:
gizmo_the_mogwai [7]3 years ago
6 0

Answer:

Covalent bonding and non-covalent bonding

You might be interested in
How many protons, electrons, and neutrons does an atom with atomic number 50 and mass number 125 contain?
kherson [118]
The number of protons in an atom is equal the atomic number (= 50)

The sum of the Neutrons and Protons is the Atomic Mass (=125)
Neutrons + Protons=125

Plug in for the protons (50)
Neutrons +(50)=125

Then, once solved, we have Neutrons = 75

Assuming the atom is NOT an ion, the amount of electrons is equal to the number of protons. (protons=electrons=50=50)

Therefore:
There are 50 electrons, 50 protons, and 75 electrons.

5 0
3 years ago
The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for
tester [92]

The given question is incomplete. The complete question is:

The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:

A. K(s)+O_2(g)\rightarrow KO_2(s)

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g)

Answer: A. K(s)+O_2(g)\rightarrow KO_2(s) : decreases

B. CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2(g) : decreases

C. CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g): no change

D. N_2O_2(g)\rightarrow 2NO(g)+O_2(g) : increases

Explanation:

Entropy is defined as the randomness of the system.

Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.

In reaction K(s)+O_2(g)\rightarrow KO_2(s), as gaseous reactant is changed to solid product, entropy decreases.

In reaction CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g), as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.

In reaction CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g), as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.

In reaction  N_2O_2(g)\rightarrow 2NO(g)+O_2(g) , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.

7 0
3 years ago
What is the product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes?
zheka24 [161]

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

Sodium borohydride is a relatively selective reducing agent  Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.

The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

to learn more about sodium borohydride and ethanol click here ,

brainly.com/question/12955502

#SPJ4

4 0
2 years ago
The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of ther
timama [110]

Answer:

Silver, 0.239 J/(g °C)

Explanation:

  • The heat change is related to specific heat as given by the formula;

Heat change = mass of substance × specific heat × change in temperature

  • Therefore; considering same amount of substance or equal masses and have the same initial temperature.
  • The change in temperature will be inversely proportional to the specific heat.
  • Therefore; the higher the specific heat lower the temperature change.
  • Hence,  the change in temperature will be highest for the substance with the lowest specific heat.

Therefore; the one that will increase in temperature the most is Silver

7 0
3 years ago
Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c
IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
3 years ago
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