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3 years ago

Why do latitudes close to the equator have warm climates?

Chemistry

1 answer:

sveticcg [70]3 years ago

6 0

Answer:

Because the equator is closer to the sun. ... Because the sun's rays hit the earth's surface at a higher angle at the equator. e. Because the sun is always directly overhead at the equator.

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Draw one line connecting each factor that increases the rate of a

VLD [36.1K]

Answer:

Adding a catalyst - More collisions every second and more collisions with enough energy to break bonds.

Increase in pressure - more collisions every second

Increase in temperature - more collisions every second with enough energy to break bonds

Explanation:

According to the collision theory, chemical reaction occurs as a result of collision between reacting particles. Only particles that possess energy above the activation energy of the reaction can collide and result in product formation. Collision of particles having energy less than the activation energy merely result in elastic collisions.

Adding a catalyst lowers the activation energy of the reaction. If the activation energy is lowered, more reactants collide and more of those collisions now have enough energy to break bonds.

When the temperature is increased, the particles become more energetic hence more collisions with energy to break bonds occur.

Increase in pressure brings the reactant particles into close proximity hence more collisions occur.

4 0

3 years ago

A common chemicals used as a rising agent

Len [333]

I believe the correct answer is 84.007 g/mol hope this helps

8 0

2 years ago

What are the different acid bases

Alex

Answer:

different acids include acetic acid,lactic acid,tartaric acids DNA(deoxyribonucliec acid) .

different bases include calcium oxide Potassium hydroxide ,Sodium hydroxide and Barium hydroxide .

5 0

3 years ago

ASAP PLEASEE What is the electron configuration of Cl using the noble gas method?

jenyasd209 [6]

Answer:

It would be this below.

Explanation:

I am not 100% sure this is write but I did my best this is still kinda hard for me. I just did this a few years ago and i struggled with it but I have been doing better since then.

Hope this helps :))

4 0

2 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

6 0

3 years ago