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3 years ago

PLEASE HELP 20 points

Chemistry

1 answer:

777dan777 [17]3 years ago

4 0

n = 1.5atm (15L) / .0821 (280k) = .98 mol NaCl

NaCl = 22.99g Na + 35.45g Cl = 58.44g NaCl

58.44g NaCl x .98 mol NaCl = 57.27g NaCl

Explanation:

hope you get it right :)

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A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

deff fn [24]

Explanation:

Entropy means the amount of randomness present within the molecules of the body of a substance.

Relation between entropy and microstate is as follows.

S = $K_{b} \times ln \Omega$

where, S = entropy

$K_{b}$ = Boltzmann constant

$\Omega$ = number of microstates

This equation only holds good when the system is neither losing or gaining energy. And, in the given situation we assume that the system is neither gaining or losing energy.

Also, let us assume that $\Omega$ = 1, and $\Omega'$ = 0.833

Therefore, change in entropy will be calculated as follows.

$\Delta S = K_{b} \times ln \Omega' - K_{b} \times ln \Omega$

= $1.38 \times 10^{-23} \times ln(0.833) - 1.38 \times 10^{-23} \times \times ln(1)$

= $1.38 \times 10^{-23} \times (-0.182)$

= $-0.251 \times 10^{-23}$

or, = $-2.51 \times 10^{-24}$

Thus, we can conclude that the entropy change for a particle in the given system is $-2.51 \times 10^{-24}$ J/K particle.

8 0

3 years ago

How many moles of electrons is required to deposit 5.6g of iron from a solution of iron (2) tetraoxosulphate(6)

yuradex [85]

Answer:

0.20 mol

Explanation:

Let's consider the reduction of iron from an aqueous solution of iron (II).

Fe²⁺ + 2 e⁻ ⇒ Fe

The molar mass of Fe is 55.85 g/mol. The moles corresponding to 5.6 g of Fe are:

5.6 g × 1 mol/55.85 g = 0.10 mol

2 moles of electrons are required to deposit 1 mole of Fe. The moles of electrons required to deposit 0.10 moles of Fe are

0.10 mol Fe × 2 mol e⁻/1 mol Fe = 0.20 mol e⁻

7 0

3 years ago

PH + pOH = 14. What is the hydroxide concentration of a solution with pH of 8?

sladkih [1.3K]

PH + pOH = 14
pOH=14-8=4
(OH-)=10^-4

4 0

4 years ago

The labels have fallen off three bottles containing powdered samples of metals; one contains zinc, one lead, and the other plati

Alinara [238K]

Here we have to identify the metal powder by the given disposal.

The identification of Zinc can be done by 1 m nitric acid (HNO₃) and Ni(NO₃)₂ which will produce hydrogen gas by reaction and displacement reaction as shown below.

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

The identification of lead can be done by the reaction with 1 m nitric acid (HNO₃) which produces lead nitrate.

The reaction is-

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

The identification of platinum can be done by the reaction with all the given disposal as it will not react with any of the compound.

1. Identification of Zinc (Zn):

(a) Zn metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than zinc.

Zn + NaNO₃ = No reaction

(b) Zn metal will react with 1 m HNO₃ to form hydrogen gas. The reaction is:

Zn + 2 HNO₃ = Zn(NO₃)₂ + H₂ (g)

(c) Zn will react with nickel nitrate [Ni(NO₃)₂] because it may only cause displacement reaction the reduction potential of Zn²⁺/Zn (-0.76) is less than that of Ni²⁺/Ni (-0.23). Thus the reaction will be:

Zn + Ni(NO₃)₂ = Zn(NO₃)₂ + Ni

2. Identification of lead (Pb):

(a) Pb metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pb.

Pb + NaNO₃ = No reaction

(b) Pb reacts with HNO₃ to form lead nitrate. The reaction is:

Pb + 4HNO₃ = Pb(NO₃)₂ + 2NO₂ + 2H₂O

(c) The standard reduction potential of Pb²⁺/Pb is more than nickel Ni²⁺/Ni thus there will be no reaction between Pb and NI(NO₃)₂.

Pb + Ni(NO₃)₂ = No reaction.

3. Identification of platinum (Pt)

(a) Pt metal will not react with sodium nitrate (NaNO₃) as sodium remains at the lower position of the activity series than Pt.

Pt + NaNO₃ = No reaction.

(b) The standard reduction potential of Pt²⁺/Pt is so high (+1.188) thus there will be no reaction with HNO₃.

Pt + HNO₃ = No reaction

(c) The standard reduction potential of Pt²⁺/Pt is more than nickel Ni²⁺/Ni thus there will be no reaction between Pt and Ni(NO₃)₂.

Pt + Ni(NO₃)₂ = No reaction.

8 0

3 years ago

Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.

bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

For the reaction:

2 HgO(s) → 2 Hg(l) + O₂(g)

ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = 181,66 kJ/mol

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

ΔSr= 216,8 J/Kmol = 0,216 kJ/Kmol

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

5 0

3 years ago