There are two reasons that account for the change in concentration. One is by the physical change and second is by the chemical change. The physical change refers to the mixture of the two compounds. The concentration is the amount of that certain substance to the whole mixture. Since the mixture increases because you add them up, the concentration decreases. Second, these compounds react with each other once they interact. Hence, their amounts would be consumed in order to produce the products.
Answer:
50g
Explanation:
At the end of the burning process, the mass of the jar and smoke should be 50g after the candle burning is done.
In most systems, the law of conservation of matter finds expression. According to this law "matter is neither created nor destroyed but atoms are rearranged".
In this kind of system, no matter is allowed to escaped and the starting mass of the reactants compared to that of the product will be the same.
Osmium is the densest<span> element on earth. It is a hard, weak, pale blue white move metal in the platinum aggregate that is found as a follow component in amalgams, generally in platinum minerals. Osmium is the densest normally happening component, with a thickness of22.59 g/cm3. There is still some open deliberation that Iridium is the denser component as per hypothetical figurings yet starting at now, Osmium has the most noteworthy measured thickness. </span>
Answer:
One object has more mass than the other object
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Answer:
The answer to your question is below
Explanation:
Data
Volume = 1000 ml
Concentration = 2M
molecule = NaCl
Process
1.- Calculate the number of moles of NaCl
Molarity = moles/Volume
-Solve for volume
moles = Molarity x Volume (liters)
-Substitution
moles = 2 x 1
-Result
moles = 2
2.- Determine the molar mass of NaCl
NaCl = 23 + 35.5 = 58.5 g
3.- Calculate the mass of NaCl to prepare the solution
58.5 g ----------------- 1mol
x ----------------- 2 moles
x = (2 x 58.5) / 1
x = 117g
4.- Weight 117 g of NaCl, place them in a volumetric flask (1 l), and add enough water to prepare the solution.
