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Gekata [30.6K]
3 years ago
6

PLEEEASSEEEE HELPPPPP!!!!!Take some time to research a utility plant. If there is one in your area, you may even visit it.

Chemistry
1 answer:
raketka [301]3 years ago
6 0

Answer:

There are four laws of thermodynamics that define fundamental physical quantities (temperature, energy, and entropy) and that characterize thermodynamic systems at thermal equilibrium.

Explanation:

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Which is NOT a type of fracture?
Andrews [41]
B. Perfect ---------
7 0
3 years ago
Name four types of pathogens that could cause infectious disease??
Kryger [21]
Bacteria, virus, fungi, and protists are the four pathogens.
3 0
3 years ago
To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water
Masteriza [31]

Answer:Sample Absorbance (625 nm)  

A 0.536  

B 0.783  

C 0.045  

Therefore, I will use these data to solve your question. If you have other absorbances values, just follow my steps and plug in different numbers.

First, we see 1 mole of NH3 gives 1 mole product.

In B moles of NH3 = moles of NH3 in A + (5.5 x10^-4 x2.5/1000) = 1.375 x10^6 + mA

( mA = moles of NH3 in A) vol of B = 25 = vol of A

now A = el C = eC ( since l = 1cm)

Because, n net absorbance due to complex blank absorbance must be removed.

Here A(A) = 0.536 - 0.045 = 0.491 , A(B) = 0.783 - 0.045 = 0.738  

(you can plug in different numbers in this step)

A2/A1 = C2/C1 , A(B)/A(A) = (1.375x10^-6 +mA)/(mA) = 0.738/0.491

So, mA = 2.733 x 10^-6 = moles of NH3 in A (Lake water)

Hence [NH3] water ( 2.733 x10^-6 ) x 1000/25 = 1.093 x 10^-4 M

Lake water vol = 10 ml out of 25,

Concentration of ammonia in lake water = 2.733 x10^-6 x 1000/10 = 2.733 x 10^-4 M

Then, A = 0.491 = e x 1 x 1.093 x10^-4

e = 4492 M-1cm-1

Explanation:

4 0
3 years ago
How long does it take the sun to melt a block of ice at 0∘c with a flat horizontal area 1.0 m2 and thickness 1.8 cm ? assume tha
VLD [36.1K]
This is a problem involving heat transfer through radiation. The solution to this problem would be to use the formula for heat flux.

ΔQ/Δt = (1000 W/m²)∈Acosθ

A is the total surface area:
A = (1 m²) + 4(1.8 cm)(1m/100 cm)(√(1 m²))
A = 1.072 m²

ΔQ is the heat of melting ice.
ΔQ = mΔHfus
Let's find its mass knowing that the density of ice is 916.7 kg/m³.
ΔQ = (916.7 kg/m³)(1 m²)(1.8 cm)(1m/100 cm)(<span>333,550 J/kg)
</span>ΔQ = 5,503,780 J

5,503,780 J/Δt = (1000 W/m²)(0.05)(1.072 m²)(cos 33°)
<em>Δt = 122,434.691 s or 34 hours</em>

4 0
3 years ago
Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the energy of a single photon of this r
denis23 [38]

Answer :

(1) The frequency of photon is, 3\times 10^{10}Hz

(2) The energy of a single photon of this radiation is 1.988\times 10^{-23}J/photon

(3) The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

Explanation : Given,

Wavelength of photon = 1.0cm=0.01m     (1 m = 100 cm)

(1) Now we have to calculate the frequency of photon.

Formula used :

\nu=\frac{c}{\lambda}

where,

\nu = frequency of photon

\lambda = wavelength of photon

c = speed of light = 3\times 10^8m/s

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

\nu=\frac{3\times 10^8m/s}{0.01m}

\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz    (1Hz=1s^{-1})

The frequency of photon is, 3\times 10^{10}Hz

(2) Now we have to calculate the energy of photon.

Formula used :

E=h\times \nu

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

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

E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})

E=1.988\times 10^{-23}J/photon

The energy of a single photon of this radiation is 1.988\times 10^{-23}J/photon

(3) Now we have to calculate the energy in J/mol.

E=1.988\times 10^{-23}J/photon

E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)

E=11.97J/mol

The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

3 0
3 years ago
Read 2 more answers
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