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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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How many grams (of mass m) of glucose are in 285 ml of a 5.50% (m/v) glucose solution?
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The percentage of glucose given is m/v. This means that the given percentage of volume consists of mass.
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3 years ago
At 25 ∘ C , the equilibrium partial pressures for the reaction 3 A ( g ) + 4 B ( g ) − ⇀ ↽ − 2 C ( g ) + 3 D ( g ) were found to
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<u>Answer:</u> The standard Gibbs free energy of the given reaction is 6.84 kJ

<u>Explanation:</u>

For the given chemical equation:

3A(g)+4B(g)\rightleftharpoons 2C(g)+3D(g)

The expression of K_p for above equation follows:

K_p=\frac{(p_C)^2\times (p_D)^3}{(p_A)^3\times (p_B)^4}

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Putting values in above expression, we get:

K_p=\frac{(4.22)^2\times (5.52)^3}{(5.70)^3\times (4.00)^4}\\\\K_p=0.0632

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy = ?

R = Gas constant = 8.314 J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_{eq} = equilibrium constant at 25°C = 0.0632

Putting values in above equation, we get:

\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (0.0632)\\\\\Delta G^o=6841.7J=6.84kJ

Hence, the standard Gibbs free energy of the given reaction is 6.84 kJ

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How is the mass in grams of the element converted to amount in atoms
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