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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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Diano4ka-milaya [45]

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Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is C_1_4H_1_8N_2O_5 and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})

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3 0
3 years ago
above shows a balloon full of gas which has a volume of 120.0ml at 300.0k assuming pressure remains constant , what is the voume
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128 ml  is the voume of the balloon if the temperature of the gas increases to 320.0k.

Explanation:

given that:

T1 (initial temperature) = 300K

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From the equation

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Since pressure is constant the equation will be:

\frac{V1}{T1} = \frac{V2}{T2}

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4 0
3 years ago
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c = specific heat of metal= ?

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c=\frac{186.75 J}{19 g\times (175^oC-35^oC)}

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Explanation:

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Answer:

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