Answer:
The elements in the reactants are the same as the elements in the products.
Explanation:
You can't have more or less elements or atoms, but you also can't just have the same number of atoms because then it could be different elements, which can't happen.
Particles below the surface of a liquid
Answer:
The reaction of an acid and a base is called a neutralization reaction. ... However, in the reaction between HCl(aq) and Mg(OH) 2(aq), additional molecules of HCl and ... First, we will write the chemical equation with the formulas of the reactants ... The chloride ions are the only spectator ions I THINKS
Explanation:
Answer:
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Explanation:
it shows it in the word search
Answer:
About 16.1 grams of oxygen gas.
Explanation:
The reaction between magnesium and oxygen can be described by the equation:

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.
Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:
- Convert grams of MgO to moles of MgO.
- Moles of MgO to moles of O₂
- And moles of O₂ to grams of O₂.
The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.
Dimensional analysis:

In conclusion, about 16.1 grams of oxygen gas was reacted.
You will obtain the same result if you compute with the 24.4 grams of Mg instead:
