The chemical reaction of magnesium and sodium hydroxide would yield magnesium hydroxide and sodium. The chemical reaction is expressed as:
Mg(s)+ 2NaOH(aq)→Mg(OH)2(s)+2Na
In ionic form,
Mg(s) + 2Na+ + 2OH−(aq)→Mg(OH)2(s) + 2Na
The hydroxide ions keep decreasing and the hydrogen ions increase, pH decreases.
<h3>What is hydroxide?</h3>
Hydroxide is a diatomic anion with synthetic recipe OH⁻. It comprises of an oxygen and hydrogen particle kept intact by a solitary covalent bond, and conveys a negative electric charge. It is a significant yet generally minor constituent of water. It capabilities as a base, a ligand, a nucleophile, and an impetus. Hydroxide is a diatomic anion with chemical formula OH −. It consists of an oxygen and hydrogen atom held together by a single covalent bond, and carries a negative electric charge. It is an important but usually minor constituent of water. Hydroxide ions can act as a catalyst in different types of reactions. Hydroxide ions can function as base, ligand, nucleophile or a catalyst.
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Answer:
3.375moles of O₂
Explanation:
The reaction expression is given as;
2KClO₃ → 2KCl + 3O₂
Number of moles of KClO₃ = 2.25moles
Now;
To find the number of moles of O₂ produced we use known number of moles.
So;
2 mole of KClO₃ will produce 3 moles of O₂
2.25moles of KClO₃ will produce 
= 3.375moles of O₂
Dalton gathers evidence for the existence of atoms by measuring the masses of elements after compounds are formed.
<u>Explanation</u>:
- John Dalton accumulated proof for the presence of atoms by estimating the majority of components that responded to frame mixes. All components are made out of molecules. All particles of a similar component have a similar mass, and atoms of various components have various masses. Mixes contain atoms of more than one component.
-
Dalton did numerous investigations that gave proof to the presence of particles. For instance, He researched pressure and different properties of gases, from which he induced that gases must comprise of little, singular particles that are in steady, arbitrary movement.
Answer:
0.800 mol
Explanation:
We have the amounts of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with moles of the compounds involved.
Step 1. <em>Gather all the information</em> in one place.
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
n/mol: 4.00 4.00
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Step 2. Identify the <em>limiting reactant
</em>
Calculate the <em>moles of CO₂</em> we can obtain from each reactant.
<em>From C₃H₈:</em>
The molar ratio of CO₂: C₃H₈ is 3:1
Moles of CO₂ = 4.00 × 3/1
Moles of CO₂ = 12.0 mol CO₂
<em>From O₂</em>:
The molar ratio of CO₂: O₂ is 3:5.
Moles of CO₂ = 4.00 × ⅗
Moles of CO₂ = 2.40 mol CO₂
O₂ is the limiting reactant because it gives the smaller amount of CO₂.
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Step 3. Calculate the <em>moles of C₃H₈ consumed</em>.
The molar ratio of C₃H₈:O₂ is 1:5.
Moles of C₃H₈ = 4.00 × ⅕
Moles of C₃H₈ = 0.800 mol C₃H₈
