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Elenna [48]
3 years ago
10

A 1.803 g sample of gypsum, a hydrated salt of calcium sulfate, CaSO4 is heated at a temperature greater than 170 degree Celsius

in a crucible until a constant mass is reached. The mass of anhydrous CaSO4 salt is 1.426g. Calculate the percent by mass of water in the hydrated calcium sulfate salt.
Chemistry
1 answer:
brilliants [131]3 years ago
6 0

Answer:

20.91% of the hydrated calcium sulfate salt is water

Explanation:

<u>Step1</u> : Calculate the mass of H2O that evaporated

A hydrated salt has a mass of 1.803g. After heating at 170° C, this means water will evaporate, the mass of the salt is 1.426g

The difference is water that evaporated.

⇒Mass H2O of hydration =1.803 g - 1.426g g = 0.377 g H2O

<u>Step 2</u>: Calculate the % of water in the hydrated calcium sulfate salt

⇒ mass of H2O = 0.377g

⇒ mass of calcium sulfate salt = 1.803g

⇒ % water in the calcium sulfate salt = (0.377 / 1.803) x 100% = 20.91%

20.91% of the hydrated calcium sulfate salt is water

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Similar propoties is the answer

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad
ASHA 777 [7]

0 \; \textdegree{\text{C}}

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

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Let the final temperature of the system be t \; \textdegree{\text{C}}. Thus \Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial})  - t_{0} = t \; \textdegree{\text{C}}

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The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable t.

E(\text{absorbed} ) = E(\text{released})

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Confirm the uniformity of units, equate the two expressions and solve for t:

66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)

t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}} which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., t \le 0 \; \textdegree{\text{C}}. However, there's no way for the temperature of the system to go below 0 \; \textdegree{\text{C}}; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at 0 \; \textdegree{\text{C}}; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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The various pressures are measured by precision instruments called barometers or pressure sensors and expressed in inches of mercury or millibars. <span>Air has a weight too, although not very much, If you "pile" the air mile high, the bottom pressure is heavier because of all the air sitting on top of it, therefore the pressure decreases with altitude, because there is less air "piled up" </span>
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