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vagabundo [1.1K]
3 years ago
5

A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution before the addition of any lioh

.
Chemistry
1 answer:
snow_lady [41]3 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

 <span>moles HClO4 = 0.100 L x 0.18 M = 0.018 
moles LiOH = 0.030 L x 0.27 = 0.0081 
moles H+ in excess = 0.018 - 0.0081 = 0.0099 
total volume = 0.130 L 
[H+] = 0.0099/ 0.130= 0.0762 M 
pH = 1.12</span>
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Give the name of the reaction that achieves the theoretical limits with respect to free energy in thermodynamics.a. reversible r
Assoli18 [71]

Answer:

The correct option is: a. reversible reaction

Explanation:

In thermodynamics, Gibb's free energy is the quantitative measure of the <u>spontaneity or feasibility </u>of a chemical reaction, at fixed temperature and pressure.

It can also be described as the <u>maximum available work obtained from a closed system</u>. This maximum work can only be achieved in a reversible process, <u>at fixed pressure and temperature.</u>

<u>The Gibb's free energy (ΔG) is given by</u>: ΔG = ΔH - T.ΔS

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3 years ago
The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium co
Anarel [89]

Answer:

Kp = 1.41 x 10⁻⁶

Explanation:

We have the chemical equation:

2 A(g) + 3 B(g)⇌ C(g)

In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):

dn= (sum moles products - sum moles reactants)

   = (moles C - (moles A + moles B))

   = (1 - (2+3))

   = 1 - 5

   = -4

We have also the following data:

Kc = 63.2

T= 81∘C + 273 = 354 K

R = 0.082 L.atm/K.mol (it is a constant)

Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:

Kc = (RT)^{dn}= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶

3 0
3 years ago
When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

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c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

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m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

3 0
3 years ago
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Answer:

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Time taken = \frac{Number of grains to be distributed}{grains distributed in 1 year}

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3 years ago
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Answer:

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Explanation:

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