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rosijanka [135]
2 years ago
8

A student used 1.2 g of hydrate in the first trial and 1.6 g in the second trial. Explain fully why her mass percent of water wi

ll be the same for both trials even though she used more hydrate in the second trial. (Do not simply state that the mass percent is an intensive property ).
Chemistry
1 answer:
Pachacha [2.7K]2 years ago
8 0

Both trials of 1.2 g and 1.6 g will have the same mass percent of water because the ratio of the salt to the water of hydration is always constant for any hydrated salt.

<h3>Water of hydration</h3>

For every hydrated salt, the ratio of the salt to the water of hydration remains constant irrespective of the amount of salt taken for experimental analysis.

For example, assuming the mass percent of water in 10g of a hydrated salt is 40%, if 100g of the same salt is taken, the mass percent will remain 40%.

More on water of hydration can be found here: brainly.com/question/11202174

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Answer:

25.7 kJ/mol

Explanation:

There are two heats involved.

heat of solution of NH₄NO₃ + heat from water = 0

q₁  +  q₂  =  0

n  =  moles of NH₄NO₃  =  8.00 g NH₄NO₃  ×  1 mol NH₄NO₃/80.0 g NH₄NO₃          

∴ n =   0.100 mol NH₄NO₃

q₁ = n * ΔHsoln = 0.100 mol * ΔHsoln

m  =  mass of solution  =  1000.0 g + 8.00 g  =  1008.0 g

q₂  =  mcΔT  = 58.0 g  ×  4.184 J°C⁻¹  g⁻¹  × ((20.39-21)°C) = -2570.19 J

q₁  +  q₂  =  0.100 mol  ×ΔHsoln  – 2570.19 J  =  0

ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

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What is the pOH of a solution that has an H+=7.7x10^-7 M
pentagon [3]

Answer:

Explanation:

Step 1

The question is based on the concept of PH and pOH calculations.

pH is defined as negative logarithmic of hydronium Ion concentration.

while pOH is defined as negative logarithmic of hydroxide ion concentration of the solution.

Step 2

[H+] = 7.7*10-7  M  

pH = -log[H+]  

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