Anyone help plz And ty

Give the half equation to show what happens to oxygen when it is rusting

ycow [4]

Answer:

½O 2 + 2e - + H 2O → 2OH.

Explanation:

Redox reactions - Higher

In terms of electrons:

oxidation is loss of electrons

reduction is gain of electrons

Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:

iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-

oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-

iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-

3 0

3 years ago

If the stack of bricks on the right has a mass of 240 kg, approximately what is the mass of the stack of bricks on the left?

solong [7]

Answer:

720

Explanation: study island

4 0

2 years ago

Read 2 more answers

Is energy used during an explosion

Whitepunk [10]

Yes explosion is mainly energy

5 0

3 years ago

Read 2 more answers

True or false that thrust is a force that propels the plane in a desired direction pls helppp fast!

hjlf

Answer:

True

Explanation:

Thrust is the force which moves an aircraft through the air. Thrust is used to overcome the drag of an airplane, and to overcome the weight of a rocket. Thrust is generated by the engines of the aircraft through some kind of propulsion system.

8 0

2 years ago

g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) +

iren2701 [21]

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Mg: 1 mole
HCl: 2 moles
MgCl₂: 1 mole
H₂: 1 mole

Being the molar mass of each compound:

Mg: 24.3 g/mole
HCl: 36.45 g/mole
MgCl₂: 95.2 g/mole
H₂: 2 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Mg: 1 mole* 24.3 g/mole= 24.3 g
HCl: 2 moles* 36.45 g/mole= 72.9 g
MgCl₂: 1 mole* 95.2 g/mole= 95.2 g
H₂: 1 mole* 2 g/mole= 2 g

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

$Molarity=\frac{number of moles of solute}{volume}$

in units $\frac{moles}{liter}$

then, the number of moles of HCl that react is:

$6 M=\frac{number of moles of HCl}{0.01 L}$

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

$mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}$

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, Mg will be the limiting reagent.

7 0

3 years ago