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2 years ago

Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.

Engineering

1 answer:

marin [14]2 years ago

8 0

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61

= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

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Answer:

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is given in the attachment.

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2 years ago

The fatigue data for a brass alloy are given as follows: Stress Amplitude (MPa) Cycles to Failure 170 3.7 × 104 148 1.0 × 105 13

prohojiy [21]

Answer:

i) S–N plot is attached

ii) fatigue strength = 100 MPa

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Explanation:

i) I have attached the S–N plot (stress amplitude versus logarithm of cycles to failure)

ii) The question says we should find the fatigue strength at 4 × 10^(6) cycles.

So let's find the log of this and trace it on the graph attached.

Log(4 × 10^(6)) = 6.6

From the graph attached, at log of cycle value of 6.6, the fatigue strength is approximately 100 MPa

iii) The question says we should find the fatigue life for 120 MPa.

Thus, from the graph, at stress amplitude of 120 MPa, the log of cycles is approximately 5.75.

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2 years ago

Use the predicate specifications(x, y): x beats yF (x): x is an (American) football teamQ(x, y): x is quarterback of yL(x,y): x

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Answer:

a) ∀y∃x(Q(x, y))

b) (B(Jayhawks, W ildcats)→¬∀y(L(Jayhawks, y)))

c) ∃x(B(Wildcats, x) ∧ B(x, Jayhawks))

Explanation:

a) The statement can be rewritten as "For all football teams, there exists a quarterback" which is written in logical symbols.

b) The statement is an implication and thus have a premise and a conclusion. The premise states "Jayhawks beat the Wildcats" which is translated using B(x, y). The conclusion can be rewritten as "It is not the case that Jayhawks lose to all football teams".

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2 years ago

The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt

Step2247 [10]

Answer:

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Explanation:

Question: If the temperature of the sheet is 77C,what is the incident solar radiation on aday with Tinf= Tsurr= 16°C?

Given

Energy Equation of the Gas

αs * Gs * A + h * A * (T inf - Tg) + εσA (Tsurr⁴- Tg⁴) = 0

Where σ= 5.67 *10^-8 W/m²K⁴ (Stefan-Boltzmann constant)

ε = 0.13 (Emisivity)

αs = 0.65 (Absorptivity for solar radiation)

h = 7W/m²K⁴

Tg = 77 + 273.15K = 350.15K

T inf = 16 + 273.15 = 288.15K

T surr= T inf = 288.15

Substitute the above values in the Gas Equation, we have

0.65 * Gs * A + 7 * A * (288.15 - 350.15) + 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴) = 0

0.65 * Gs * A = - 7 * A * (288.15 - 350.15) - 0.13 * 5.67 * 10^-8 * A * (288.15⁴ - 350.15⁴)

A cancels out, so we are left with

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3 0

2 years ago

An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for t

11Alexandr11 [23.1K]

Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

$E=h/t = \frac{(1/2\pi)*6.626*10x^{-34} J.s}{20*10x^{-9} } = 5.273*10x^{-27} J = 3.29* 10^{-8} eV$ .

The corresponding uncertainty in the emitted frequency ∆v is:

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To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

λ = c/v

dλ/dv = -c/v² = -λ²/c

Therefore,

∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

= 9.20 × 10⁻¹⁵ m or 9.20 fm

The length of the photon (<em>l)</em> is

l = (light velocity) × (emission duration)

= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m

6 0

2 years ago