Which of the following are part of kinetic-molecular? Check all that apply

Which of the following correctly describes the role of scientific models in the advancement of technology? Models represent the

Nikitich [7]

Models help us better understand the phenomena

3 0

3 years ago

From the given elements Na, P , Ca ,Br which one has highest first ionization energy , ,smallest atomic size , most metallic cha

madam [21]

Answer:

highest first ionization energy - Br

smallest atomic size - Br

most metallic character - Na

Explanation:

Ionization energy of Br is higher than P because higher zeff value

Sodium is more metallic than calcium because it is able to loose electron more readily as compared to calcium because of higher electro positivity.

Atomic radius of Br is the smallest as its atomic radius is 114, P (115), Ca (197), Na (186)

5 0

3 years ago

A car moved 20 km east and 60 km west. What is it’s average velocity?

Stels [109]

Answer:

60-20=40km west because the direction Will favour whichever direction is bigger

3 0

3 years ago

Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32

Law Incorporation [45]

Answer:

For a: The standard Gibbs free energy of the reaction is -347.4 kJ

For b: The standard Gibbs free energy of the reaction is 746.91 kJ

Explanation:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$ ............(1)

For a:

The given chemical equation follows:

$2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)$

Oxidation half reaction: $Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-$ ( × 2)

Reduction half reaction: $3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)$

We are given:

$n=2\\E^o_{cell}=+1.08V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ$

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

For b:

The given chemical equation follows:

$6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

Oxidation half reaction: $Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-$ ( × 6)

Reduction half reaction: $2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)$

We are given:

$n=6\\E^o_{cell}=-1.29V\\F=96500$

Putting values in equation 1, we get:

$\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ$

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0

3 years ago

Think of Earth's shape. How does the shape of Earth affect the amount of heat different

diamong [38]

Answer: The Earth is an oblate spheroid which is more elongated along the equator.

Regions along the equator receives more amount of direct sunlight from the sun causing higher temperatures along the region. While places along the poles receives less amount of heat since heat is scattered in a larger area due to the shape of the Earth around the poles.

Explanation: So places along the equator experiences higher temperature due to the direct hit of sunlight. And places along the polar regions experiences cooler temperature since it receives less amount of sunlight.

8 0

3 years ago