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4 years ago

The arrows in the image show the direction in which carbon is moving Which process of the carbon cycle is indicated

1 answer:

8 0

The carbon cycle involves the circulation of carbon dioxide (CO2) from the atmosphere into plants and other living organisms; the transfer of carbon from these organisms into other temporary storage pools, living or nonliving, containing organic and inorganic carbon compounds; and the return of CO2 to the atmosphere

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Charge q is 1 unit of distance away from the source charge S. Charge p is six times further away. The force exerted between S an

QveST [7]

Answer:

36 times

Explantaion:

The force exerted between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In the case of S and q: The distance of separation is 1 unit

F(Sq) α S*q/1²

In the case of S and p: the distance is 6 units

F(Sp) α S*p/6²

therefore:

F(Sq) = 36 * F(Sp)

8 0

3 years ago

Select True or False: All intermolecular forces must be overcome in order for a substance to undergo a phase change from a liqui

lozanna [386]

Answer:

True

Explanation:

In order for a liquid to boil it must undergo transformation in its structural composition that are arranged by intermolecular. Such as changes will need more energy to overcome and changes in phases form solid to liquid to gas.

4 0

3 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Fiesta28 [93]

This is an incomplete question, here is a complete question.

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

$M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)$

Substance ΔG°f (kJ/mol)

M₃O₄ -9.50

M(s) 0

O₂(g) 0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

$PCl_3(l)\rightarrow PCl_3(g)$

First we have to calculate the Gibbs energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]$

where,

$\Delta G^o$ = Gibbs energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]$

$\Delta G^o=9.50kJ/mol$

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant = ?

$9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)$

$K=0.0216$

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

$K=(P_{O_2})^2$

$0.0216=(P_{O_2})^2$

$P_{O_2}=0.147atm$

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0

4 years ago

From this pics I just took... how old do you think I am..?

Kruka [31]

Answer:

uhhhhhhh like what 13 or twelve

Explanation:

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3 years ago

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Part 1: Fill in the blank Newton’s Second Law: Unbalanced forces cause an object to ______.

blagie [28]

Answer:

and can cause changes in motion. Inertia.

Explanation:

4 0

3 years ago

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