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marusya05 [52]
2 years ago
15

Label A-F based on the table using C for concentrated and D for dilute. A 3-column table with 3 rows. Column 1 is labeled Acid s

lash base with entries upper H c l, Upper N a upper O upper H, Upper H subscript 2 upper S upper O subscript 4. Column 2 is labeled Molarity with entries 12 M and 0.5 M, 0.01 M and 6.0 M, 0.05 M and 10 M. Column 3 is labeled concentrated slash dilute with entries A and B, C and D, E and F.
Chemistry
2 answers:
Agata [3.3K]2 years ago
8 0

Answer : C , D , D , C , D , C

stellarik [79]2 years ago
3 0

Answer:

A) C

B) D

C) D

D) C

E) D

F) C

Explanation:

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Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab
uranmaximum [27]

The answer is B) 230 m3

5 0
2 years ago
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The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

Now put all the given values in this expression, we get:

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

Now put all the given values in this expression, we get

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia
tiny-mole [99]

Answer:

The percent yield of this reaction is 70%

Explanation:

The reaction is: N₂ + 3H₂ → 2NH₃

We only have the mass of H₂, so we assume that N₂ is in excess

We convert the mass to moles, to work with the reaction:

450 g . 1mol / 2 g = 225 moles

Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia

Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles

This is the 100% yield reaction → We convert the moles of NH₃ to mass

150 mol . 17g /1mol = 2550 g

Percent yield = (Produced yield/Theoretical yield) .100

Percent yield = (1575g/2550g) . 100 = 70%

7 0
2 years ago
Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f
Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0
2 years ago
Which of the following type of protons are chemically equivalent? A) homotopic B) enantiotopic C) diastereotopic A &amp; B B &am
Rina8888 [55]

Answer:

A) homotopic and B) enantiotopic

Explanation:

Protons chemically equivalent are those that have the same chemical shift, also if they are interchangeable by some symmetry operation or by a rapid chemical process.

The existence of symmetry axes, Cn, that relate to the protons results in the protons being homotopic, that is chemically equivalent in both chiral and aquiral environments.

The existence of a plane of symmetry, σ, makes the protons related by it, are enantiotopic and these protons will only be equivalent in an aquiral medium; if the medium is chiral both protons will be chemically NOT equivalent. The existence of a center of symmetry, i, in the molecule makes the related protons through it enantiotopic and therefore chemically only in the aquiral medium.

Diastereotopic protons cannot be interconverted by any symmetry operation and they are different, with different chemical displacement.

6 0
2 years ago
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