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Masja [62]
3 years ago
14

Which of the following equations does not demonstrate the law of conservation of mass

Chemistry
2 answers:
tia_tia [17]3 years ago
8 0
The answer is c :) i hope its the right answer
Alex_Xolod [135]3 years ago
5 0

Answer:

The answer is A

Explanation:

Just took a test.

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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
victus00 [196]

Explanation:

The given data is as follows.

        q = 69.0 kJ = 69000 J (as 1 kJ = 1000 J),

       mass (m) = 8.10 kg = 8100 g  (as 1 kg = 1000 g)

        T_{i} = 33.9^{o}C = (33.9 + 273) K = 306.9 K

         C = 4.18 J/gK

As we know that the relation between heat and change in temperature is as follows.

                   q = m \times C \times \Delta T

Putting the values into the above formula to calculate the final temperature as follows.

                  q = m \times C \times \Delta T

       69000 J = 8100 g \times 4.18 J/g K \times (T_{f} - 306.9 K)                

           69000 J = 33858 \times (T_{f} - 306.9 K)

                  (T_{f} - 306.9 K) = 2.037 K

                    T_{f} = (2.037 + 306.9) K

                                  = 308.9 K

or,                               = 309 K (approx)

Thus, we can conclude that the new temperature of the water bath is 309 K.

5 0
3 years ago
Describe the relationship between the mole and molar mass. Be prepared to describe the process for converting between moles and
malfutka [58]

Answer:

The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. ... In such a conversion, we use the molar mass of a substance as a conversion factor to convert mole units into mass units (or, conversely, mass units into mole units).

Explanation:

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2 years ago
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Climax (the most exciting or plot changing event)
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if this is true or false its true hope i helped

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3 years ago
Sodium carbonate (na2co3) is used to neutralize the sulfuric acid spill. how many kilograms of sodium carbonate must be added to
jenyasd209 [6]
<span>5.45 x 10^3 kg of sodium carbonate is needed to neutralize 5.04 kg of sulfuric acid. For this, I will assume you have pure H2SO4. So first, you need to calculate the molar mass of H2SO4 and Na2CO3. Lookup the atomic weights of all the elements involved. Atomic weight of Sodium = 22.989769 Atomic weight of Sulfur = 32.065 Atomic weight of Carbon = 12.0107 Atomic weight of Oxygen = 15.999 Atomic weight of Hydrogen = 1.00794 Molar mass of H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass of Na2CO3 = 2 * 22.989769 + 12.0107 + 3 * 15.999 = 105.987238 g/mol The balanced equation for the reaction of Na2CO3 with H2SO4 is Na2CO3 + H2SO4 => Na2SO4 + CO2 + H2O so for every mole of sulfuric acid to be neutralized, you need 1 mole of sodium carbonate. You can determine the number of moles of sulfuric acid you have and then calculate the mass of that many moles of sodium carbonate. But, there's an easier way. Just use the relative mass differences between sodium carbonate and sulfuric acid. So 105.987238 g/mol / 98.07688 g/mol = 1.080655 So that means for every kg of sulfuric acid, you need 1.080655 kg of sodium carbonate. Now do the multiplication. 5.04 x 10^3 kg * 1.080655 = 5.4465 x 10^3 kg. Since you only have 3 significant figures for your data, round the result to 3 significant figures, giving 5.45 x 10^3 kg</span>
3 0
3 years ago
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