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padilas [110]
3 years ago
9

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Br2(l) and I-(aq) Use the reduction po

tential values for Br2(l) of 1.07 V and for I2(s) of 0.53 V Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.). Do NOT include spaces, units, punctuation or anything else silly!
Chemistry
1 answer:
Vikki [24]3 years ago
3 0

Answer:

1.58×10E18

Explanation:

Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.

E°cell= 1.07- 0.53= 0.54 V

E°cell= 0.0592/n logK

0.54 = 0.0592/2 logK

logK= 0.54/0.0296

logK= 18.2

K= Antilog (18.2)

K= 1.58×10^18

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Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou
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39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

Ca(s) + Cl_2(g) → CaCl_2(s)

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

m_{ca_C_l_2}= 16.7 g Ca x \frac{1 mol \;of \;Ca}{40g Ca} x \frac{1 mol \;of \;CaCl_2}{1 mol \;Ca} x \frac{111g of \;CaCl_2}{1 mol \;CaCl_2}

m_{ca_C_l_2} = 39.96 g

Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

Learn more about the stoichiometric process here:

brainly.com/question/15047541

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All atoms and molecules have London Dispersion Forces between them, but they are usually overshadowed but the much stronger forces. In this scenario the major attractive force in HF molecules are hydrogen bonds. Hydrogen bonds are electrostatic forces of attraction found when Hydrogen is bonded to a more electronegative atom such as Oxygen, Chlorine and Fluorine.

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