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san4es73 [151]
3 years ago
6

How many grams of CaCl2 are need to prepare 250 mL of a 0.15M solution?

Chemistry
1 answer:
Tanzania [10]3 years ago
4 0

Answer:

Here is what I got from reading this question...

Explanation:

So if you dissolve 7.5625- g of CaCl2 in water, dissolve well and then add distilled in a 250- ml standard flask upto the mark and shake well for uniform Concentration, you will get 0.25- M CaCl2 solution.

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A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate
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<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of solution 1 (liquid water) = 50.0 g

m_2 = mass of solution 2 (liquid water) = 29.0 g

T_{final} = final temperature = ?

T_1 = initial temperature of solution 1 = 25°C  = [273 + 25] = 298 K

T_2 = initial temperature of solution 2 = 45°C  = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K

Converting this into degree Celsius, we use the conversion factor:

T(K)=T(^oC)+273

305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC

Hence, the final temperature of water is 32.3°C

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