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Roman55 [17]
3 years ago
14

Enter a balanced equation for the reaction between aqueous lead(II)(II) nitrate and aqueous sodium chloride to form solid lead(I

I)(II) chloride and aqueous sodium nitrate. Express your answer as a chemical equation. Identify all of the phases in your answer. Pb(NO3)2(aq)+2NaI(aq)→PbI2(s)+2NaNO3(aq) You have already submitted this answer. Enter a new answer. No credit lost. Try again.
Chemistry
1 answer:
aev [14]3 years ago
3 0

Answer:

Pb(NO₃)₂ (aq)  + 2NaCl (aq)  → PbCl₂ (s) ↓  + 2NaNO₃ (aq)

Explanation:

The reactants are:

Lead(II) nitrate → Pb(NO₃)₂ (aq)

Sodium chloride → NaCl (aq)

The products are:

Lead(II) chloride → PbCl₂ (s)

Sodium nitrate → NaNO₃ (aq)

Salts form nitrate are soluble. The chloride makes a precipitate with the Pb²⁺. The chemical equation for this reaction is:

Pb(NO₃)₂ (aq)  + 2NaCl (aq)  → PbCl₂ (s) ↓  + 2NaNO₃ (aq)

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Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms
nirvana33 [79]

Answer:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }

With further simplification, we have:

P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]

Then, we have:

P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]

Therefore,

PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]

Using the expansion:

(1-x)^{-1} = 1 + x + x^{2} + ....

Therefore,

PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]

Thus:

PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]           equation (1)

Using the virial equation of state:

P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]

Thus:

PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]     equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol[/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

3 0
2 years ago
The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?
spin [16.1K]
The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] >  [H(+)] * [F(-)] the option a. fo the list of choices.
8 0
3 years ago
Read 2 more answers
Select the pair that has the larger atom or ion listed first.
Sedbober [7]

Answer:

Correc option: Br^- \, , Kr

Explanation:

size of atom : it says somthing about how many shell present in a particular atom or ion and it can also be evaluated on the basis of radius of atom.

Br^- and Kr has highest number of shell as compared to other group of species .

Na ,S , Mg ,P all are from 3rd period but Kr and Br^- in the 4th period so size of species of this group will more,

Size increases on increasring the shell number

3 0
3 years ago
How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?
mars1129 [50]

Answer:

111.15 g are required to prepare 500 ml of a 3 M solution

Explanation:

In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.

Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of

(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.

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Which highly reactive gas was probably absent from the Earth's primitive atmosphere? O2 (oxygen gas) water vapor methane carbon
rosijanka [135]
Oxygen gas was most likely absent from Earth's primitive atmosphere.  The current theory is that the Earth's early atmosphere was composed of mainly carbon dioxide and methane due to the high volcanic activity.  Cyanobacteria and their use of photosynthesis was what caused earth's atmosphere to become oxygen enriched.
I hope that helps.
6 0
3 years ago
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