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3 years ago

When a metal reacts with oxygen gas what is the product formed

Chemistry

2 answers:

beks73 [17]3 years ago

6 0

Answer:

Hello There!!

Explanation:

When a metal reacts with oxygen gas metal oxide forms.Whenever a compound or element reacts with oxygen It turns to oxide.

hope this helps,have a great day!!

~Pinky~

Fittoniya [83]3 years ago

6 0

Answer:

Metal + Oxygen -> Metal Oxide

Explanation:

It's pretty self-explanatory. However, let me show you an example:

Iron + Oxygen -> Iron Oxide

Iron rusts. When iron rusts, it comes in contact with air to form Iron Oxide, which is rust itself.

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Balance the following skeleton ionic reaction, and calculate E°cell to decide whether the reaction is spontaneous: Ag(s) + Cu2+

tia_tia [17]

Answer:

2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)

Explanation:

Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.

E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V

The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.

5 0

3 years ago

Read 2 more answers

In the laboratory you are asked to make a 0.565 m sodium bromide solution using 315 grams of water. How many grams of sodium bro

Scorpion4ik [409]

Answer : The mass of sodium bromide added should be, 18.3 grams.

Explanation :

Molality : It is defined as the number of moles of solute present in kilograms of solvent.

Formula used :

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}$

Solute is, NaBr and solvent is, water.

Given:

Molality of NaBr = 0.565 mol/kg

Molar mass of NaBr = 103 g/mole

Mass of water = 315 g

Now put all the given values in the above formula, we get:

$0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}$

$\text{Mass of NaBr}=18.3g$

Thus, the mass of sodium bromide added should be, 18.3 grams.

6 0

3 years ago

As the temperature of a liquid goes up, the average speed of its particles .This means that the kinetic energy of the particles

kirill115 [55]

The kinectic energy would be decreasing.

7 0

3 years ago

What is the definition of an atmosphere? a portion of the electromagnetic spectrum that makes the sky look blue the blanket of g

goldenfox [79]

Answer:

the blanket of gases that surrounds Earth and some other planets

Explanation:

The best definition of the atmosphere is that it is a portion of the earth that is a made up of a blanket of gases.

The atmosphere is not peculiar to the earth alone, some other planets also have atmospheric cover.
The earth's atmosphere is divided into many different layers.
The weather we see and experience on the earth surface is conditioned within the troposphere.
This troposphere is the closest layer to the surface.

7 0

2 years ago

Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th

stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH= $2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}$

Knowing:

ΔH= 75.5 kJ/mol
$H_{NO}$ = 90.25 kJ/mol
$H_{Cl_{2} }$ = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
$H_{NOCl}$ =?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - $H_{NOCl}$

Solving

- $H_{NOCl}$ =75.5 kJ/mol - 2*90.25 kJ/mol

- $H_{NOCl}$ =-105 kJ/mol

$H_{NOCl}$ =105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0

2 years ago