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Papessa [141]
3 years ago
9

2) __H,PO, +

Chemistry
1 answer:
Gnoma [55]3 years ago
4 0

Explanation:

a.H2PO4 + 2KOH = K2PO4 + 2H2O

b. H2PO4 _ 2KOH

1(2) + 31 + 16(4) _ 2(23) + 2(16) + 2(1)

2 + 31 + 64 _ 46 + 32 + 2

97 _ 80

no of moles of H2PO4 = 2.5 ÷ 97

= 0.026mol

no of moles of 2KOH = 4 ÷ 80

= 0.05mol

2KOH is the limiting reactant

c. H2PO4 is the excess reactant

d. 1 mole H2PO4 weighs 2.5g

0.026mol H2PO4 weighs g

(0.026 × 2.5)

H2PO4 weighed 0.065g after reaction

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nekit [7.7K]

Answer:

A and D are true , while B and F statements are false.

Explanation:

A) True.  Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

thus the standard gibbs free energy can be determined using Keq

B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions

C) False. From the equation presented

ΔG⁰ = (-RT*ln Keq)

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T= 298 K

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5 0
3 years ago
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hram777 [196]

Answer:

no

Explanation:

7 0
2 years ago
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What do butterflies and lobsters have in common?
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Explanation:

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3 years ago
Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a
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Answer:

                    SEPARATION SCHEME FOR  CATIONS

GIVEN  CATIONS : Ag^{+} \ ,  Fe^{3+} , Cu^{2+}, Ni^{2+}

     

    Step 1:   Add 6mol/dm^3 of HCl to the mixture solution

    Result : This would cause a precipitate of AgCl to be formed

    Reaction :  Ag^{+} _{(aq)} + Cl^{-} _{(aq)}  ---------> AgCl(ppt)

    Step 2 : Next is to remove the precipitate and add H_2S to the remaining          

                 solution in the presence of 0.2 \ mol/dm^3 of HCl

     Result : This would cause a precipitate of CuS to be formed

     Reaction :  Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)

 

     Step 3: Next remove the precipitate then add 6 \ mol/dm^3 of aqueous      

                 NH_3 (NH_3 \cdot H_2 O) , process the solution in a centrifuge,when the  

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                 Now this precipitate is   Fe(OH)_3 and the remaining solution

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                 Next take out the precipitate to a different beaker and add HCl

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Explanation:

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