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stepan [7]
3 years ago
14

Use the solubility rules from the Lab 4 introduction and your knowledge of qualitative separation schemes from the lab to answer

the following questions. The qualitative analysis experiment you did is actually an abbreviated version of a much larger analysis scheme in which many different cations are separated and identified. Suppose a mixture contains Ag , K , NH4 , Hg22 , Pb2 , Mg2 , Sr2 , Ba2 , Cu2 , Al3 and Fe3 .
(a) Which of the following ions could you separate, by causing them to precipitate, with the addition of HCl?
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
(b) After the addition of HCl, the above sample is centrifuged and decanted. Which of the following cations remaining in the supernatant could you separate, by causing them to precipitate, with the addition of H2SO4? (Hint: H2SO4 is a source of sulfate ions. Select all that apply.)
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
Chemistry
1 answer:
cluponka [151]3 years ago
4 0

Answer:

a13+a13

Explanation:

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Aluminum chloride can be formed from its elements:
saul85 [17]

<u>Answer:</u> The \Deltas H^o_{formation} for the reaction is -1406.8 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 reaction for the formation reaction of AlCl_3 is:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3 (s)    \Delta H^o_{formation}=?

The intermediate balanced chemical reaction are:

(1) HCl(g)\rightarrow HCl(aq.)    \Delta H_1=-74.8kJ    ( ×  6)

(2) H_2(g)+Cl_2(g)\rightarrow 2HCl(g)    \Delta H_2=-185kJ     ( ×  3)

(3) AlCl_3(aq.)\rightarrow AlCl_3(s)    \Delta H_3=+323kJ     ( ×  2)

(4) 2Al(s)+6HCl(aq.)\rightarrow 2AlCl_3(aq.)+3H_2(g)    \Delta H_4=-1049kJ

The expression for enthalpy of formation of AlCl_3 is,

\Delta H^o_{formation}=[6\times \Delta H_1]+[3\times \Delta H_2]+[2\times \Delta H_3]+[1\times \Delta H_4]

Putting values in above equation, we get:

\Delta H^o_{formation}=[(-74.8\times 6)+(-185\times 3)+(323\times 2)+(-1049\times 1)]=-1406.8kJ

Hence, the \Deltas H^o_{formation} for the reaction is -1406.8 kJ.

6 0
3 years ago
Does the mass of water increase or decrease when it changes to ice
NikAS [45]

Answer:

The mass stays the same only volume changes, the volume decreases

Explanation:

The ice shrinks (decreases volume) and becomes more dense. The weight will not (and cannot) change.

4 0
3 years ago
Read 2 more answers
CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of
neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0
3 years ago
A solution is saturated in CO2 gas and KNO3 at room temperature. What happens when the solution is warmed to 75°C?
nignag [31]

Answer:

gaseous CO2 bubbles out of the solution

Explanation:

We already know that the dissolution of a gas in water is exothermic. Hence, when the temperature of a solution containing a gas is increased, the solubility of the gas decreases and the gas bubbles out of the solution.

Similarly, the dissolution of KNO3 in water is endothermic. This implies that the solubility of the solid increases with increasing temperature.

Thus the solid becomes more soluble at 75°.

6 0
3 years ago
Give the number of significant figures indicated 6.695
Genrish500 [490]
3 significant figures 

5 0
3 years ago
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