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VARVARA [1.3K]
3 years ago
14

How many milliliters of 0.0510 m edta are required to react with 50.0 ml of 0.0200 m cu2 ?

Chemistry
1 answer:
krok68 [10]3 years ago
4 0
Cu  ions plus EDTA2-   ->cu(EDTA)2-  plus 2H-
   number  of  moles  of  CU  ions  used  which is   equal to  molarity  multiplied  by    volume  in  litres
that  is  50xo.o2  divided  by 1000  that  is  0.001moles
Since  ratio  is  1:1  the  moles  of EDTA  is  also0.001moles
volume  of  EDTA  is  0.001  divided  by  0.0510m  which  is  0.0196  litres in  ml is  0.0196x1000  which  is  19.61ml
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Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g) + 2H2(g) ⇌ C2H6(
sukhopar [10]

Answer:

-255.4 kJ

Explanation:

The free energy of a reversible reaction can be calculated by:

ΔG = (ΔG° + RTlnQ)*n

Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.

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Q = 0.261/[8.58*(3.06)²]

Q = 3.2487x10⁻³

ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)

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4 years ago
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Answer:

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kolbaska11 [484]

Answer:

Answers are in the explanation

Explanation:

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Now, the KOH reacts with HF, thus:

KOH + HF →  F⁻ + H₂O

<em>That means after reaction, concentration of HF decrease increasing F⁻ concentration.</em>

Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.

For the statements:

A. The number of moles of HF will increase. <em>FALSE</em>. HF react with KOH, thus, moles of HF decrease

B. The number of moles of F- will decrease. <em>FALSE</em>. The reaction produce  F⁻ increasing its moles.

C. The equilibrium concentration of H₃O⁺ will increase. <em>FALSE. </em>The equilibrium shift to the left decreasing concentration of H₃O⁺

D. The pH will decrease. <em>FALSE</em>. As the H₃O⁺ concentration decrease, pH will increase

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