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2 years ago

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Be sure to answer all parts. calculate δg o for the reaction between i2(s) and br−(aq). e o cell = −0.54 j/c enter your answer i

n scientific notation. δg o = × 10 j

1 answer:

denpristay [2]2 years ago

3 0

Answer:

See explaination

Explanation:

See attachment for the detailed step by step solution of the given problem.

The attached file have the solved problem.

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Convert 338 L at 63.0 atm to its new volume at standard pressure.

taurus [48]

The new volume at standard pressure of 1 atm is 21294 liters.

Explanation:

Data given:

Initial volume of the gas V1 = 338 liters

initial pressure on the gas P1 = 63 atm

standard pressure as P2 = 1 atm

Final volume at standard pressure V2 =?

The data given shows that Boyle's law equation is to used:

P1V1 = P2V2

rearranging the equation to calculate V2,

V2 = $\frac{P1V1}{P2}$

Putting the values in the equation:

V2 = $\frac{338X63}{1}$

= 21294 L

as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.

7 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago

In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Ke

kondor19780726 [428]

Answer:

Explanation:

Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.

In the living E. coli cells,

[ATP] = 7.9 mM;

[ADP] = 1.04 mM,

[glucose] = 2 mM,

[glucose 6-phosphate] = 1 mM.

Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.

The reaction is given as

Glucose + ATP → glucose 6-phosphate + ADP

Now reaction quotient for given equation above is

$q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}$

$q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}$

so,

$q$ ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq

4 0

2 years ago

1. A student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drinks. The student

Lina20 [59]

Answer:

1) The bubbles will grow, and more may appear.

2)Can A will make a louder and stronger fizz than can B.

Explanation:

When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.

Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.

3 0

2 years ago

In which 1.0 gram sample are particles arranged in a rystal structure?

lilavasa [31]

The Options are as follow,

<span> (1) CaCl</span>₂<span> (s) (3) CH</span>₃<span>OH (l)</span>

<span> (2) C</span>₂<span>H</span>₆<span> (g) (4) Cal</span>₂<span> (aq)</span>

Answer:

Option-1 is the correct answer.

Explanation:

As we know crystal formation is the property of solids. Therefore, in given options we are given with four different states of matter.

Option A, CaCl₂ is in a solid state , so it can exist in crystal form.

Option 2, C₂H₆ (Ethane) is in gas form, so it cannot form crystals.

Option 3, CH₃OH (Methanol) is present in liquid form, so it fails to form crystals.

Option 4, CaI₂, it is dissolved in water, Hence, it is in aqueous state, Therefore it also lacks crystal structure.

5 0

2 years ago