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3 years ago

A prairie dog eats grass. A coyote eats the prairie dog. This is an example

Chemistry

1 answer:

Oduvanchick [21]3 years ago

5 0

Answer:

This is an example of a food chain

Explanation:

Think of it as a chain reaction. The grass feeds and nourishes the prairie dog. Upon eating the prairie dog, the coyote gets the nutrients from both the grass the prairie dog ate and from the prairie dog itself.

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Compare and contrast single-bonded compounds and double-bonded compounds.

max2010maxim [7]

The main difference between single double and triple bond is the number of shared electrons. If the shared number is one pair of electrons, the bond will be a single bond, whereas if two atoms bonded by two pairs (four electrons), it will form a double bond.

Souce: pediaa.com/difference-between-single-double-and-triple-bonds/

3 0

4 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

PLS HELP ME ASAP!!!!!

vichka [17]

Independent variable- Sally's hair

Dependent variable- The shininess of Sally's hair

Control- The Shampoo

7 0

3 years ago

Give the name of the following molecule. EXTRA POINTS PLEASE HELP

Rudik [331]

There isn t a 'middle' carbon because there is an even number of them

If the methyl group braches off the 3rd carbon the name is

hexa-3-methyl-3-ene

3 0

3 years ago

How many grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel?

Nimfa-mama [501]

Answer: 0.87 g of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 5.5% w/w gel.

Explanation:

According to the dilution law,

$C_1w_1+C_2w_2=C_3w_3$