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2 years ago

Of ozone, molecular oxygen, and oxygen radicals, which is the most stable form of oxygen? How do you know?

1 answer:

5 0

Oxygen has a strong double bond which has more stability than the single co-ordinate bond in ozone, therefore more energy is required to break the O2 bonding than ozone, so the ozone molecule is more reactive than oxygen gas. ... The oxygen free radical contains two unpaired electrons in its valence shell.

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Need help with question please ..<br> -20 points

san4es73 [151]

Answer:

B oklolol

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8 0

2 years ago

If the radius of a circle is 3.5 inches, its diameter is inches and its approximate area is square inches.

Monica [59]

Answer: Diameter = 7 inches

Area of circle is 38.46 inches²

Explanation: The diameter is the radius x 2 3.5 x 2 = 7 inches .

Hope this helps!!! Plz mark me!!!! Thx :)

8 0

3 years ago

Read 2 more answers

The sulfur and nitrogen compounds in smog combine with water to form__.

erastova [34]

The sulfur and nitrogen compounds in smog combine with water to form Acid Rain. Acid rain is harmful to the environment and infrastructure because of its aidic components.

7 0

3 years ago

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A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

3 years ago

Which of the following atoms has the weakest attraction for electrons in a bond with a calcium (Ca) atom?

Paladinen [302]

Answer:

B- Iodine

Explanation:

Using the periodic trends, electronegativity (electron affinity) goes down as you go down a period. This is due to the atom gaining new shells when going down each period, resulting in a weakened attraction force between the nucleus and the electrons in the cloud. In other words. as more shells are added, the weaker the atom's affinity for electrons are and the less electronegative it is. As a result, Iodine would have the weakest attraction for electrons in a bond with Calcium as it is the furthest down in the periodic table.

6 0

3 years ago