All categories

3 years ago

What is the most common type of climate/ecosystem found in the Congo River basin?

Chemistry

2 answers:

Sedbober [7]3 years ago

6 0

Answer:

It is the Tropical Rainforest.

Explanation:

I took the test.

I hope this was helpful!

avanturin [10]3 years ago

4 0

The answer is tropical rainforest. The congo river basis is a mosaic of rivers, forests, savannas, swamps and flooded forests.

You might be interested in

Which models are most appropriate for him to use?

jekas [21]

Answer:

the triangular model and the cubic model

5 0

3 years ago

Read 2 more answers

1. What is one of the important roles of tropical rainforests in the earth's environment?

Makovka662 [10]

Answer:

1.D

2.C

3.D

4.D

5.C

6 B

those are the answers, we must take care of our environment

3 0

3 years ago

How can you demonstrate waves?

Lady bird [3.3K]

Shaking a phone cord, strumming a guitar string, playing a trumpet

7 0

3 years ago

Of the choices below, which is true for the relationship shown? it is ka for the acid h3p2o72â. it is kb for the acid h3p2o72â.

Semenov [28]

A For PLATO. just did it

3 0

3 years ago

Read 2 more answers

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

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

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago