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marshall27 [118]
3 years ago
5

H2O (s) ⇒ H2O What would be the best argument against writing this in the form of a chemical equation?

Chemistry
1 answer:
Goshia [24]3 years ago
5 0

Answer: Since this is merely a Physical Change, it is improper to write it as a chemical equation.

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On which energy level are the valence electrons for calcium (Ca) ?
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The fourth (last) one in 2-8-8-2.
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2 years ago
1. What amount of ammonia (in moles) is produced by the reaction of 4.00 mol H2 with 3.00 mol Nz?
ArbitrLikvidat [17]
<h2>Answer:  6 moles</h2>

<h3>Explanation:</h3>

3 H₂    +    N₂   →   2 NH₃

   ↓            ↓

4 mol       3 mol

Since the moles of N₂ is the smaller of the two reactants, then N₂ is the limiting factor (the reactant that will decide how much ammonia is produced since it has the smaller amount of moles). ∴ we have to use it in calculating the number of moles of ammonia

The mole ratio of N₂ to NH₃ based on the balanced equation is 1 to 2.

∴ the moles of NH₃ = moles of N₂ × 2

                                =  3 moles × 2

                                = 6 moles

7 0
2 years ago
When 0.250 moles of KCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change is observed. Given
777dan777 [17]

Explanation:

Upon dissolution of KCl heat is generated and temperature of the solution raises.

Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

             17.24 kJ/mol \times 0.25 mol

                = 4.31 kJ

or,             = 4310 J      (as 1 kJ = 1000 J)

Mass of solution will be the sum of mass of water and mass of KCl.

       Mass of Solution = mass of water + (no. of moles of KCl × molar mass)

                                    = 200 g + (0.25 mol \times 54.5 g/mol)

                                    = 200 g + 13.625 g

                                    = 213.625 g

Relation between heat, mass and change in temperature is as follows.

                             Q = mC \Delta T

where,    C = specific heat of water = 4.184 J/g^{o}C

Therefore, putting the given values into the above formula as follows.

                     Q = mC \Delta T

            4310 J = 213.625 g \times 4.184 J/g^{o}C \times \Delta T      

              \Delta T = 4.82^{o}C

Thus, we can conclude that rise in temperature will be 4.82^{o}C.

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3 years ago
Calculate the pH of the following <br> A.0.02mol/L HCI<br>B. 0.1 mol/L NaOH​
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