QUE NOMBRE RECIBE LA MASA GASEOSA QUE ENVUELVE LA TIERRA

Oxygen is composed of only one kind of atom and can’t be separated into simpler substances. What type of matter is it? A. soluti

drek231 [11]

The answer is B. element.

An element is composed of only one kind of atom and cannot be separated into simpler substances. Oxygen (O) is the element.
A compound is a substance composed of two or more different atoms chemically bonded to one another, for example, water (H₂O) consists of 2 atoms of hydrogen (H) and 1 atom of oxygen (O), so it is the compound.
A mixture consists of two or more substances that are not chemically combined. Solutions and colloids are mixtures.

8 0

2 years ago

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How do the fusion reaction in the sun compare to the fusion occurring in large atars and supernovas

r-ruslan [8.4K]

The fusion reaction in the sun is a combination of hydrogen atoms fusing to create helium. The fusion reaction in larger stars involve much heavier elements like oxygen and iron. In supernovas, often elements like gold are produced

6 0

2 years ago

The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate

Cloud [144]

Answer : The $\Delta G$ for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $2Ag^++2e^-\rightarrow 2Ag$

Now we have to calculate the Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o$

where,

$\Delta G^o$ = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole$

Therefore, the $\Delta G$ for this reaction is, -88780 J/mole.

7 0

3 years ago

Nitrogen monoxide and water react to form ammonia and oxygen, like this: 4no (g) +6h2o (g) →4nh3 (g) +5o2 (g) the reaction is en

Archy [21]

The concept used here is the Le Chatelier's principle. When a disturbance is introduced to the system, it favors the direction of reaction that minimizes the disturbance to regain equilibrium.

In endothermic reactions, the forward reaction is favored when the temperature is low. Otherwise, the reverse reaction is favored. When you add the amounts of substances on the reactant side, more products would formed favoring the forward reaction. If you increase concentration on the product side, you form more reactants so it would favor the reverse reaction. Lastly, since 10 moles of gases are needed in the reactant side, it would be favored during high pressure reaction.

3 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

3 years ago