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2 years ago

Why do smokers have a more difficult time getting rid of bronchitis

Chemistry

2 answers:

Digiron [165]2 years ago

6 0

Answer:

C. Smoking damages the cilia in the respiratory system

son4ous [18]2 years ago

4 0

Answer:

C. Smoking damages the cilia in the respiratory system.

Explanation:

Hope this helped.

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The track along which electricity flows is called a: fuse circuit switch

Pavel [41]

It is called a circuit.

4 0

2 years ago

wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.

Papessa [141]

Answer: The value of equilibrium constant for the net reaction is 11.37

Explanation:

The given chemical equations follows:

Equation 1: $A+2B\xrightarrow[]{K_1} 2C$

Equation 2: $2C\xrightarrow[]{K_2} D$

The net equation follows:

$D\xrightarrow[]{K} A+2B$

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K=K_1\times \frac{1}{K_2}$

We are given:

$K_1=2.57$

$K_2=0.226$

Putting values in above equation, we get:

$K=2.57\times \frac{1}{0.226}=11.37$

Hence, the value of equilibrium constant for the net reaction is 11.37

6 0

3 years ago

The ionization energies for removing successive electrons from sodium are 496 kJ/mol, 4562 kJ/mol,

ddd [48]

Answer:

Explanation:

The high jump of ionization energy indicates that we are trying to remove electron from noble gas configuration state.

The ionization energy data specifies that the Elements are from group 1 at period 3 or greater.

Removing the first electron require 496 kJ and the second ionization energy jump significantly due to the removal of electron from the noble gas configuration which is logical because electron try to maintain the especially stable state.

3 0

3 years ago

How many milligrams of AgNO3 is required to completely react with 81.5 mg LiOH?

BartSMP [9]

Answer:

$m_{AgNO_3}=577.6mg$