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2 years ago

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A 3-m3 rigid tank contains hydrogen at 250 kPa and 550 K. The gas is now cooled until its temperature drops to 350 K. Determine

(a) the final pressure in the tank and (b) the amount of heat transfer.

1 answer:

san4es73 [151]2 years ago

6 0

Answer:

A. 159.1 kPa.

B. 686.23 kJ.

Explanation:

Below is an attachment containing the solution.

You might be interested in

For the reaction: MgF2(s) ⇌ Mg2+(aq) + 2F- (aq), Ksp= 6.4 × 10-9, the addition of 0.10 M NaF to the solution cause what effect o

galina1969 [7]

Answer:

Shifts the equilibrium to the left. reduces solubility.

Explanation:

MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

S S 2S

∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²

⇒ 4S² * S = 6.4 E-9

⇒ 4S³ = 6.4 E-9

⇒ S³ = 1.6 E-9

⇒ S = 1.1696 E-3 M

NaF(s) → Na+(aq) + F-(aq)

0.10M 0.10M 0.10M

MgF2(s) ↔ Mg2+(aq) + 2F-(aq)

S' S' 2S' + 0.10

⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²

If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:

⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'

⇒ S' = 6.4 E-7 M

∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption

We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).

8 0

2 years ago

Write a hypothesis for this: Hummingbirds are attracted to the color red

Margaret [11]

Answer: I believe/don't believe hummingbirds are attracted to the color red because ... (Enter your reasons)

Explanation:

Hypothesis are always different, up to you to choose believe or not believe depending on your position.

:) Hope this helped.

8 0

2 years ago

Bismuth(III) sulfate is an ionic compound formed from Bi3+ and SO42-. What is the correct way to represent the formula?

Hitman42 [59]

Answer:

Bi2(SO4)3

Explanation:

Bismuth(iii) sulfate is an ionic compound therefore, their is transfer of electron. Ionic compound has both cations and anions. The cations is positively charged ion while the anions is negatively charged ions. The cations loses electron to become positively charged while the anions gains electron to become negatively charged.

From the compound above, Bismuth(iii) sulfate the cations will be Bismuth ion which loses 3 electrons. The anions is the sulfate ion (S04)2- with a -2 charge.

The chemical formula can be computed from the charge configuration as follows

Bi3+ and (SO4)2-

cross multiply the charges living the sign behind to get the chemical formula

Bi2(SO4)3

Note the final chemical formula, the numbers are sub scripted

4 0

2 years ago

The standard molar heat of fusion of ice is 6020 j/mol. calculate q, w, and ∆e for melting 1.00 mol of ice at 0◦c and 1.00 atm p

zysi [14]

Answer : q = 6020 J, w = -6020 J, Δe = 0

Solution : Given,

Molar heat of fusion of ice = 6020 J/mole

Number of moles = 1 mole

Pressure = 1 atm

Molar heat of fusion : It is defined as the amount of energy required to melt 1 mole of a substance at its melting point. There is no temperature change.

The relation between heat and molar heat of fusion is,

$q=\Delta H_{fusion}(\frac{Mass}{\text{ Molar mass}})$ (in terms of mass)

or, $q=\Delta H_{fusion}\times Moles$ (in terms of moles)

Now we have to calculate the value of q.

$q=6020J/mole\times 1Mole=6020J$

When temperature is constant then the system behaves isothermally and Δe is a temperature dependent variable.

So, the value of $\Delta e=0$

Now we have to calculate the value of w.

Formula used : $\Delta e=q+w$

where, q is heat required, w is work done and $\Delta e$ is internal energy.

Now put all the given values in above formula, we get

$0=6020J+w$

w = -6020 J

Therefore, q = 6020 J, w = -6020 J, Δe = 0

3 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