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BlackZzzverrR [31]
3 years ago
10

QUESTION 3 Consider a solution containing 0.80 M NaF and 0.80 M HF. Calculate the moles of HF and the concentration of HF after

addition of 20.0 mL of 0.40 M HCl to 60.0 mL of this buffer solution. 0.088 moles, 0.99 M 0.055 moles, 0.11 M 0.048 moles, 0.60 M 0.088 moles, 0.20 M 0.056 moles, 0.93 M 0.040 moles, 0.67 M 0.056 moles, 0.70 M
Chemistry
1 answer:
Lisa [10]3 years ago
6 0

Answer:

0.056moles HF and 0.70M

Explanation:

When a strong acid is added to a buffer, the acid reacts with the conjugate base.

In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:

NaF + HCl → HF + NaCl

Initial moles of NaF and HF in 60.0mL of solution are:

NaF:

0.0600L × (0.80mol / L)= 0.048 moles NaF

HF:

0.0600L × (0.80mol / L)= 0.048 moles HF

Then, the added moles of HCl are:

0.0200L × (0.40mol / L) = 0.008 moles HCl.

Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:

<em>0.056moles HF</em>

<em></em>

In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:

0.056mol HF / 0.0800L = <em>0.70M</em>

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If I add 475mL of water to 75.0mL of a 0.315M NaOH solution, what will the molarity of the diluted solution be?
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Answer is: <span>the molarity of the diluted solution 0,043 M.
</span>V(NaOH) = 75 mL ÷ 1000 mL/L = 0,075 L.
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n(NaOH) = c(NaOH) · V(NaOH).
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balandron [24]
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4 0
2 years ago
Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with
Sphinxa [80]

Answer : The work done is, 1.98\times 10^4J

Explanation :

The given balanced chemical reaction is:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

When 4 moles of N_2 react with 12 moles of H_2 then it gives 8 moles of NH_3

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of N_2 + Moles of H_2

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of NH_3

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas = 25^oC=273+25=298K

Putting values in above equation, we get:

1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K

V=195.7L

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

w=-p\Delta V

where,

w = work done

p = pressure of the gas = 1.0 atm

\Delta V = change in volume = -195.7 L

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

w=-p\Delta V

w=-(1.0atm)\times (-195.7L)

w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, 1.98\times 10^4J

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Note that:

Particles in a:
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solid are tightly packed, usually in a regular pattern.
Particles in a:
gas vibrate and move freely at high speeds.
liquid vibrate, move about, and slide past each other.
solid vibrate (jiggle) but generally do not move from place to place.
Liquids and solids are often referred to as condensed phases because the particles are very close together.
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Some Characteristics of Gases, Liquids and Solids and the Microscopic Explanation for the Behavior
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little free space between particles
flows easily
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