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Juli2301 [7.4K]
3 years ago
10

Consider the equations below. (1) Ca(s) + CO2(g) + 1 2 O2(g) → CaCO3(s) (2) 2Ca(s)+O2(g) → 2CaO(s) How should you manipulate the

se equations so that they produce the equation below when added? Check all that apply. CaO(s) + CO2(g) → CaCO3(s) reverse the direction of equation (2) multiply equation (1) by 3 multiply equation (2) by 1/2
Chemistry
1 answer:
In-s [12.5K]3 years ago
3 0

Answer : The correct options are, reverse the direction of equation (2) and multiply equation (2) by 1/2.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation.

The main given balanced chemical reaction is,

CaO(s)+CO_2(g)\rightarrow CaCO_3(s)

Now we have to determine the main chemical reaction from the given two intermediate reactions.

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)

(2) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)

Now reverse the equation (2), multiply equation (2) by 1/2 and then add both the equations, we get the main chemical reaction.

(1) Ca(s)+CO_2(g)+\frac{1}{2}O_2(g)\rightarrow CaCO_3(s)

(2) CaO(s)\rightarrow Ca(s)+\frac{1}{2}O_2(g)

Now add both the equations, we get:

CaO(s)+CO_2(g)\rightarrow CaCO_3(s)

Hence, the steps used for the main reaction are, reverse the direction of equation (2) and multiply equation (2) by 1/2.

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A 2.2 M solution is made by with 0.45 moles of a solute. What is the final volume of this solution?
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A compound containing chromium, Cr; chlorine, Cl; and oxygen, O, is analyzed and found to be 33.6% chromium, 45.8% chlorine, and
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Answer

The empirical formula is CrO₂Cl₂

Explanation:

Empirical formula is the simplest whole number ratio of an atom present in a compound.

The compound contain, Chromium=33.6%

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                                          Oxygen=20.6%

And the molar mass of Chromium(Cr)=51.996 g mol.

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                 Oxygen containing molar mass (O)=15.999  g mol.

Step-1

 Then,we will get,

Cr=\frac{1}{51.996} \times33.6=0.64 mol

Cl= \frac{1}{35.45} \times45.8=1.29 mol.

O=\frac{1}{15.99} \times=1.28 mol.

Step-2

Divide the mole value with the smallest number of mole, we will get,

Cr= \frac{0.64}{0.64} =1

Cl= \frac{1.29}{0.64} =2

O= \frac{1.28}{0.64} =2

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)

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