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labwork [276]
3 years ago
6

Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO.

Chemistry
1 answer:
Nadya [2.5K]3 years ago
6 0

The empirical formula is K₂CO₃.  

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.  

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of K:C:O</em>.  

I like to summarize the calculations in a table.  

<u>Element</u> <u>Moles</u>  <u>Ratio</u>¹ <u>Integers</u>²  

     K       0.104   2.00         2

     C       0.052  1.00          1

     O      0.156   3.00         3

¹ To get the molar ratio, you divide each number of moles by the smallest number.  

² Round off the number in the ratio to integers to integers (2, 1, and 3).

The empirical formula is K₂CO₃.

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A buffer solution is made using a weak acid, HA. If the pH of the buffer is 9 and the ratio of A– to HA is 100, what is the pKa
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To calculate the pKa of the weak acid, we use the Henderson-Hasselbalch equation. It is expressed as pH = pKa - log [HA]/[A-]. This equation takes into account the concentration of the substance that does not dissociates into ions since it is a weak acid. We caculate as follows:

pH = pKa - log [HA]/[A-]
9 = pKa - log 1/100
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3 0
3 years ago
Carbon, hydrogen and ethane each burn exothermically in an excess of air. AHⓇ =-393.7 kJ mol. C(s) + O2(g) → CO2(g) H2(g) + % O2
Salsk061 [2.6K]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 51.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The chemical equation for the reaction of carbon and water follows:

2C(s)+2H_2(g)\rightarrow C_2H_4(g) \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_1=-393.7kJ    ( × 2)

(2) H_2+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_2=-285.9kJ     ( × 2)

(3) 2C_2H_4(s)+2O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)    \Delta H_3=-1411kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[2\times \Delta H_1]+[2\times \Delta H_2]+[1\times (-\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(2\times (-393.7))+(2\times (-285.9))+(1\times -(-1411))]=51.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is 51.8 kJ.

6 0
3 years ago
How Do Glow Sticks Glow?
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Answer:

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Explanation:

8 0
2 years ago
What mass of aluminum is produced by the decomposition of 5.0 kg al2o3?
Nina [5.8K]
The  mass for  of aluminum that is produced  by  the decomposition  of  5.0 Kg Al2O3 is 2647 g or 2.647  Kg

          calculation
  Write  the equation for decomposition  of Al2O3

Al2O3 = 2Al  + 3 O2

find the  moles  of  Al2O3 =  mass/molar mass

convert  5 Kg  to g   = 5 x1000 = 5000 grams
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 moles =5000 g/  102 g/mol = 49.0196 moles

by use  of mole ratio between Al2O3 to  Al  which is 1:2  the moles of Al = 49.0196 x2 =98.0392  moles


mass of  Al = moles x molar  mass

= 98.0392 moles x  27g/mol = 2647  grams  or 2647/1000 = 2.647 Kg


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