A sample of silver (106.00 g) was heated inside a dry test tube in a water bath

Question

A sample of silver (106.00 g) was heated inside a dry test tube in a water bath until the silver was at the same temperature as the water (98.6 C). The silver metal was quickly transferred to 50.0 g of room-temperature water (24.5 and allowed to transfer its excess heat to the water. At the highest temperature reached, the water and metal sample reached a temperature of 32.4 C. Find the specific heat of silver. The specific heat of water is 4.186 J/g.C. heat absorbed by the water q = mc delta t heat released by the silver = heat absorbed by water = me delta t Describe how you would determine the percentage water of hydration and the formula of the hydrate in lab.

Explanation / Answer

3. heat = mCpdT

Cp = specific heat

heat lost = heat gain

106 x Cp x (98.6 - 32.4) = 50 x 4.186 x (32.4 - 24.5)

So,

specific heat of silver Cp,

Cp = 0.2356 J/g.oC

4. Percentage water of hydration

a. Take the hydrated salt, weigh it before heating.

b. Heat the salt, weigh it after heating.

c. calculate mass loss.

d. mass lost is water of hydration.

e. calculate moles of anhydrous salt left and moles of water

f. ratio (moles of water/moles of anhydrous salt) gives number of water molecules present in the hydrated salt.

g. moles of water x 100/moles of hydrated salt is the percetage of water of hydration.

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