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andriy [413]
4 years ago
7

What precipitate forms when silver nitrate and potassium chromate solutions are mixed? Explain.

Chemistry
1 answer:
weeeeeb [17]4 years ago
3 0

Answer: a. Ag_2CrO_4( s)

Explanation:

2AgNO_3(aq)+K_2CrO_4(aq)\rightarrow Ag_2CrO_4(s)+2KNO_3(aq)

When silver nitrate AgNO_3 combines with potassium chromate K_2CrO_4(aq) ,they undergo double displacement reaction in which exchange of ions take place.

The products formed are silver chromate Ag_2CrO_4 which is insoluble in water and thus formed as precipitate and potassium nitrate KNO_3(aq)is soluble in water.


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Determine the enthalpy change for the decomposition of calcium carbonate. CaCO3 (s) --> CaO (s) + CO2 (g) given the thermoche
Gnesinka [82]

Answer : The enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol

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. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

CaCO_3(s)\rightarrow CaO(s)+CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)     \Delta H_1=65.02kJ/mol

(2) Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)    \Delta H_2=-113.8kJ/mol

(3) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_3=-393.5kJ/mol

(4) 2Ca(s)+O_2(g)\rightarrow 2CaO(s)    \Delta H_4=-1270.2kJ/mol

Now we are reversing reaction 1 and then adding reaction 1 and 2, we get :

(1) Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)     \Delta H_1=65.02kJ/mol

(2) CaCO_3(s)+H_2O(l)\rightarrow Ca(OH)_2(s)+CO_2(g)    \Delta H_2=113.8kJ/mol

The expression for enthalpy of change will be,

\Delta H=\Delta H_1+\Delta H_2

\Delta H=(65.02)+(113.08)

\Delta H=178.1kJ/mol

Thus, the enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol

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4 years ago
High-level radioactive waste requires deep burial because it is explosive. TRUE FALSE
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True.................
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What is the specific latent heat of fusion for a substance that takes 550 KJ to melt 14 KG at 262 K?
Arte-miy333 [17]

Answer:- The right choice is A. 3.9*10^4\frac{J}{kg} .

Solution:- Latent heat of fusion means the heat required to melt the solid at constant temperature means there is no change in temperature only the solid changes to liquid. So, it is a solid to liquid phase change.

q=m*\Delta H_f_u_s

where q is the heat required to convert solid to liquid, m is the mass and \Delta H_f_u_s is the latent heat of fusion.

From given info, 550 kJ that is 550000 J of heat is required to melt 14 kg of solid at 262K temperature. Let's rearrange the equation for latent heat of fusion and plug in the values in it.

\Delta H_f_u_s=\frac{q}{m}

\Delta H_f_u_s=\frac{550000J}{14kg}

\Delta H_f_u_s = 39286\frac{J}{kg}

If we round this value to two sig figs and write in scientific notations then it becomes 3.9*10^4\frac{J}{kg} .

So, the right choice is A. 3.9*10^4\frac{J}{kg} .

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Explanation:

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