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Liono4ka [1.6K]
4 years ago
7

You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a

bsorbed by the reaction being studied.
Chemistry
2 answers:
Olenka [21]4 years ago
7 0
A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT 
To find the heat released.
notka56 [123]4 years ago
3 0

Calorimeter functions by possessing a known mass of familiar substance combust or react in an enclosed space. The calorimeter exhibits an agent for captivation of the heat discharged at the time of reaction or combustion, for example, water may act as the heat absorbing agent.  

The variation in temperature of the heat absorbent in the company of its mass and specific heat capacity are utilized to find out the energy discharged with the help of the equation:  

Q = mCΔT

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Question 6 The mineral barite, (BaSO.) has a ke of 1.1 x 10" at 25°C. Calculate the solubility of barium sulfate in water, in: 6
Sav [38]

Explanation:

(6.1).    The reaction equation will be as follows.

             BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)

Assuming the value of K_{sp} as 1.1 \times 10^{-10} and let the solubility of each specie involved in this reaction is "s". The expression for K_{sp} will be as follows.

            K_{sp} = [Ba^{2+}][SO^{-}_{2}]    (Solids are nor considered)

                        = s \times s

                   s = \sqrt{K_{sp}}

                      = \sqrt{1.1 \times 10^{-10}}

                      = 1.05 \times 10^{-5}

Therefore, solubility of barium sulfate in water is 1.05 \times 10^{-5}.

(6.2).   As the molar mass of BaSO_{4} is 233.38 g/mol

Therefore, the solubility is g/L will be calculated as follows.

                233.38 g/mol \times 1.05 \times 10^{-5}

                  = 2.45 \times 10^{-3} g/L

Therefore, solubility of barium sulfate in grams per liter is 2.45 \times 10^{-3} g/L.

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3 years ago
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Answer:

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mr_godi [17]

Answer:

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From the given problem;

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3 years ago
Covalent bonding is the ____ of electrons​
taurus [48]

Answer: sharing

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