he required empirical formula based on the data provided is Na2CO3.H2O.
<h3>What is empirical formula?</h3>
The term empirical formula refers to the formula of a compound which shows the ratio of each specie present.
We have the following;
Mass of sodium = 37.07-g
Mass of carbonate = 48.39 g
Mass of water = 14.54-g
Number of moles of sodium = 37.07-g/23 g/mol = 2 moles
Number of moles of carbonate = 48.39 g/61 g/mol = 1 mole
Number of moles of water = 14.54/18 g/mol = 1 mole
The mole ratio is 2 : 1: 1
Hence, the required empirical formula is Na2CO3.H2O
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Answer:
If more solute is added and it does not dissolve, then the original solution was saturated
If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturate
Explanation:
Answer is: <span>Double Displacement.
Combustion is reaction with oxygen.
</span>Synthesis is reaction of two or more substances combining to make a more complex
substance.
Decomposition is reaction where one substance is broken down into two or more simpler substances.
Single Displacement is reaction where neutral element metal or nonmetal
become an ion as it replaces another ion in a compound.
<span>Double displacement
reactions (more reactive metals displace metals with lower reactivity).
</span>Neutralization<span>is is </span>reaction<span> in which an </span>acid<span> and a </span>base<span> react quantitatively with each other.</span>
Answer:
The new concentration is 2.03M
Explanation:
Step 1: Data given
A 200 mL 3.55 M HBr is diluted with 150 mL
Step 2: The dilution
In a dilution, the ratio that exists between the concentration of the stock solution and the concentration of the diluted solution equals the ratio that exists between the volume of the diluted solution and the volume of the stock solution.
Dilution factor = [stock sample]/[diluted sample] = diluted volume / stock volume
In this case, the volume of the stock solution is 200 mL
Adding 150 mL of water to the stock solution will dilute it to a final volume of 200 + 150 = 350 mL
The dilution factor wll be 350/200 = 1.75
This makes the diluted concentration:
3.55/1.75 = 2.03M
The new concentration is 2.03M
