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3 years ago

13

We can also use the equation for enthalpy change for physical phase changes. Consider the phase change H2O(l) → H2O(g). Calculat

e ΔHrxn.

2 answers:

BlackZzzverrR [31]3 years ago

4 0

<span>44.0 kJ is the answer</span>

Lostsunrise [7]3 years ago

3 0

The answer will be <u>44.0 kJ</u> .

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You are managing a city that needs to upgrade its disinfection basin at your 40 MGD surface water drinking water treatment plant

nignag [31]

Solution :

According to Chick's law

$\frac{N_t}{N_0}=e^{-k'C^n t}$

where, t = contact time

c = concentration of disinfectant

k' = lethality coefficient = 4.71

n = dilution coefficient = 1

4 log removal = % removal = 99.99

$\frac{N_t}{N_0}=\frac{\text{bacteria remaining}}{\text{bacteria initailly present}}$

= 1 - R

= 1 - 0.9999

Now for plug flow reactor contact time,

$\tau =\frac{V}{Q} =\frac{75000}{40 \times 10^6}$

= 0.01875 days

= 27 minutes

For CSTR, $\tau =\frac{V}{Q} =\frac{150000}{40 \times 10^6}$

$=3.75 \times 10^{-3}$ days

= 5.4 minute

There are 3 reactors, hence total contact time = 3 x 5.4

= 16.2 minute

Or $\frac{N_t}{N_0}=e^{-k'C^n t}$

or $(1-0.9999)=e^{-4.71 \times C \times t}$

∴ C x t = 1.955

For PFR, $t_1 = 27$ min

∴ C $=\frac{1.955}{27}$ = 0.072 mg/L

For CSIR, $t_2=16.2$ min

$C=\frac{1.955}{16.2} = 0.1206$ mg/L

∴ Chlorine required for PFR in kg/day

$=\frac{0.072 \times 40 \times 10^6 \times 3.785}{10^6}$ (1 gallon = 3.785 L)

= 18.25 kg/day

Therefore we should go for PFR system.

3 0

3 years ago

How many moles of KNO3 are needed to make 600 ml of a 1.3M solution?

ludmilkaskok [199]

M = n / V

Where, M is molarity (M or mol/L), n is number of moles of the solute (mol) and V is volume of the solution (L).

Here the solute is KNO₃.
The given molarity is 1.3 M
This means 1L of solution has 1.3 moles of KNO₃.

Hence moles in 600 mL = 1.3 M x 0.6 L = 0.78 mol

Therefore to make 1.3 M KNO₃ solution, needed moles of KNO₃ is 0.78 mol

7 0

3 years ago

What is the pH of 0.10 M NaF(aq). The Ka of HF is 6.8 x 10-4

dsp73

<u>Given information:</u>

Concentration of NaF = 0.10 M

Ka of HF = 6.8*10⁻⁴

<u>To determine:</u>

pH of 0.1 M NaF

<u>Explanation:</u>

NaF (aq) ↔ Na+ (aq) + F-(aq)

[Na+] = [F-] = 0.10 M

F- will then react with water in the solution as follows:

F- + H2O ↔ HF + OH-

Kb = [OH-][HF]/[F-]

Kw/Ka = [OH-][HF]/[F-]

At equilibrium: [OH-]=[HF] = x and [F-] = 0.1 - x

10⁻¹⁴/6.8*10⁻⁴ = x²/0.1-x

x = [OH-] = 1.21*10⁻⁶ M

pOH = -log[OH-] = -log[1.21*10⁻⁶] = 5.92

pH = 14 - pOH = 14-5.92 = 8.08

Ans: (b)

pH of 0.10 M NaF is 8.08

6 0

3 years ago

Classify these definitions as that of an Arrhenius acid, an Arrhenius base, or other. Arrhenius acid definition Arrhenius base d

stealth61 [152]

Answer:

Explanation:

A substance that produces an excess of hydroxide ion (-OH) in aqueous solution.

This is an arrhenius Base

According to the arrhenius theory, a base is a substance that combines with water to produce excess hydroxide ions, OH⁻ in an aqeous solution. Examples are :

Sodium hydroxide NaOH
Potassium hydroxide KOH

A substance that produces an excess of hydrogen ion (H+) in aqueous solution

This is an arrhenius Acid

An arrhenius acid is a substance that reacts with water to produce excess hydrogen ions in aqueous solutions.

Examples are;

Hydrochloric acid HCl
Hydroiodic acid HI
Hydrobromic acid HBr

5 0

3 years ago

Identify the three types of metamorphism.

Elena-2011 [213]

Answer:

Three types of metamorphism exist: contact, dynamic, and regional. Metamorphism produced with increasing pressure and temperature conditions is known as prograde metamorphism.

4 0

3 years ago