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Ronch [10]
2 years ago
10

Predict the direction in which the equilibrium will lie for the reactionC6H5COO- + HF C6H5COOH + F-.Ka(C6H5COOH) = 6.5 x 10-5; K

a(HF) = 7.1 x 10-4In the middleTo the rightTo the left
Chemistry
1 answer:
Llana [10]2 years ago
7 0

Answer:

Equilibrium will lie to the right.

Explanation:

K_{a}(HF)> K_{a}(C_{6}H_{5}COOH)

That means HF is stronger acid than C_{6}H_{5}COOH.

Alternatively, it can be said that C_{6}H_{5}COO^{-} is a stronger base than F^{-}.

Therefore HF readily gives proton to C_{6}H_{5}COO^{-} to form C_{6}H_{5}COOH and F^{-} than the reverse reaction.

So, the equilibrium will lie more towards the right i.e. towards formation of C_{6}H_{5}COOH and F^{-}.

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You can detect salt in water without tasting by measuring the density of the water. Place a glass of spring water and a glass of the suspected salt water on a balance scale and the heavier one contains salt. Other ways to test for salt in water is to put a drop of water on the end of a nail and place in a gas flame. If the water contains salt, the flame will turn a yellow/orange color.

7 0
3 years ago
Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actua
GrogVix [38]

The value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

The rust forms when 4.85X10³ kJ of heat is released is 888.916 g.

<h3>Chemical reaction:</h3>

4 Fe + 3O2 ------ 2Fe2O3

∆H = -1.65×10³kJ

A) Given,

mass of iron = 0.250kg = 250 g

<h3>Calculation of number of moles</h3>

moles = given mass/ molar mass

= 250/ 55.85 g/mol.

= 4.476 mol

As we know that,

For the rusting of 4 moles of Fe, ∆H = -1.65×10³kJ

For the rusting of 4.476 moles of Fe ∆H required can be calculated as

-1.65×10³kJ × 4.476 mol/ 4mol

∆H required = -1.846 × 10³kJ

Now,

when 2 mol of Fe2O3 formed, ∆H = - 1.65×10³kJ

It can be said that,

-1.65×10³kJ energy released when 2 mol of Fe2O3 formed

So, -4.6 × 10³kJ energy released when 2 mol of Fe2O3 formed

= 2 × -4.6 × 10³kJ / -1.65×10³kJ

= 5.57 mol of Fe2O3 formed

Now,

mass of Fe2O3 formed = 5.57 mol × 159.59 g/mol

= 888.916 g

Thus, we calculated that the rust forms when 4.85X10³ kJ of heat is released is 888.916 g. and the value of ∆H when 0.250kg of iron rusts is -1.846 × 10³kJ.

learn more about ∆H:

brainly.com/question/24170335

#SPJ4

DISCLAIMER:

The given question is incomplete. Below is the complete question

QUESTION:

Deterioration of buildings, bridges, and other structures through the rusting of iron costs millions of dollars a day. The actual process requires water, but a simplified equation is 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s) ΔH = -1.65×10³kJ

a) What is the ∆H when 0.250kg iron rusts.

(b) How much rust forms when 4.85X10³ kJ of heat is released?

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When scientists are ready to publish the results of their experiments, why is it important for them to include a description of
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Answer: To be able to not forget what have they done when they want to re-do it.

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Explanation:

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How many molecules are there in 21.4 grams of Ca(OH)2?
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