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Ivahew [28]
3 years ago
6

Titanium metal is used as a structural material in many high-tech applications such as in jet engines.

Chemistry
1 answer:
Elza [17]3 years ago
3 0

Answer:

a.Cp=0.523

b.Cp=25.02\frac{J}{mol^\circ C}

Explanation:

a. First, we solve the specific heat equation as follows:

Q=mCp\Delta T\\Cp=\frac{Q}{m\Delta T}\\Cp=\frac{89.7J}{33.0g\times5.20^\circ C}=0.523

b. Then, we use the molar mass of titanium to determine its molar heat capacity, as follows:

Cp=0.523\frac{J}{g^\circ C}\times 47.87\frac{g}{mol}\\Cp=25.02\frac{J}{mol^\circ C}

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olasank [31]
I think the Ksp for Calcium Carbonate is around 5×10⁻⁹
(I don't know if this is the Ksp value that you use because I read somewhere that this value can vary.  You should probably check with your teacher with what Ksp value they want you to use)

the equation for the dissociation CaCO₃ in water is CaCO₃(s)⇄Ca²⁺(aq)+CO₃²⁻(aq) which means that the concentration of Ca²⁺ is equal to the concentration of CO₃²⁻ in solution.  For every molecule of CaCO₃ that dissolves, one atom of Ca²⁺ and one molecule of CO₃²⁻ is put into solution which is why the concentrations are equal in solution. 

Since Ksp=[Ca²⁺][CO₃²⁻] and we know that [Ca²⁺]=[CO₃²⁻] we can rewrite the equation as Ksp=x² since if you say that [Ca²⁺]=[CO₃²⁻] when you multiply them together you get the concentration squared (I am calling the concentration x for right now).

when solving for x:
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x=0.0000707
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Using the fact that the molar mass of calcium carbonate is 100.09g/mol you can use dimensional analysis as fallows:
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That means that there is 0.007077g of Calcium carbonate that can precipitate out of 1L of water.

since the question is asking for how much water needs to be evaporated to precipitate 100mg (0.1g) of Calcium you have to do the fallowing calculation:
(0.1g)/(0.007077g/L)=14.13L of water.
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These types of questions can get long and confusing so I bolded parts that were important to try to guide you through it more easily.

I hope this helps.  Let me know if anything is unclear.
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C2H4 is oxidized and O2 is reduced in both reactions.

<h3>What is oxidation/reduction?</h3>

Oxidation is defined in several ways. Some of the definitions are:

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Reduction, on the other hand, is defined as:

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The oxidizing agent is O2. In oxidation reactions, the oxidizing agents usually get reduced. Thus, O2 is reduced in both reactions.

More on oxidation and reduction can be found here: brainly.com/question/3867774

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