Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus 
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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South Pole and north Pole . it's about megnetic field different spheres of earth interact
A reduced element (which gains electrons) and an oxidized element are required for redox reactions (gives electrons). It is not a redox reaction if we lack both of them (an element can not receive electrons if no element gives electrons and vice versa).
A reduced half and an oxidized half, which always occur together, make up redox processes. While the oxidized half experiences electron loss and an increase in oxidation number, the reduced half obtains electrons and the oxidation number declines. The mnemonic devices OIL RIG, which stand for "oxidation is loss" and "reduction is gain," are simple ways to memorize this. In a redox process, the total number of electrons stays constant. In the reduction half reaction, another species absorbs those that were released in the oxidation half reaction.
In a redox reaction, two species exchange electrons, and they are given unique names:
- The ion or molecule that accepts electrons is called the oxidizing agent - by accepting electrons it oxidizes other species.
- The ion or molecule that donates electrons is called the reducing agent - by giving electrons it reduces the other species.
Hence, what is oxidized is the reducing agent and what is reduced is the oxidizing agent.
<h3>
What is the purpose of oxidizing agents and reducing agents?</h3>
By reducing other compounds and shedding electrons, a reducing agent raises its oxidation state. An oxidizing agent gets electrons by oxidizing other compounds; as a result, its oxidation state lowers.
<h3>
What is a redox reaction?</h3>
Oxidation-reduction (or "redox") reactions are chemical processes in which electrons are exchanged between two substances. An oxidation-reduction reaction is any chemical process in which a molecule, atom, or ion alters the number of electrons it has, hence increasing or decreasing its oxidation state.
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ANSWER: B salt does not evaporate with water
