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3 years ago

Unscramble the word nlpituloo

Chemistry

2 answers:

Studentka2010 [4]3 years ago

7 0

The unscrambled word would be “Pollution”

nadezda [96]3 years ago

3 0

Hello

The word is "pollution"

Hope this helps and have a great day!

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If the pH of an acid is 4.0, what is the pOH?

LiRa [457]

Well from my research I've calculated that pH + pOH = 14.00. in conclusion
pOH = 14.0 - 4.0 and so forth that equals 10.0 your welcome

6 0

3 years ago

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Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

Which of the following statements is true based on the propositions of the kinetic molecular theory?

Delvig [45]

The answer is 4.
Gases have low densities, because of the increased space between hight-energy particles.

6 0

3 years ago

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What is produced in the electrolysis of brine?

ziro4ka [17]

High concentration of water and salt is the main ingredient of brine. Salt being NaCl and water make brain and important solution in making of chlorine.

Electric terminals are put inside the solutions and with the help of electric current the chemical properties of the solution are changed such that we get chlorine as outcome. This process is carried out in a large scale to get chlorine from NaCl in solution and is called electrolysis of Brine.

7 0

4 years ago

Approximately how many particles are in 2 moles?

schepotkina [342]

Answer:

$1.2\times10^2^4$ particles in 2 moles.

Explanation:

The number of particles that are contained in one mole, the international unit of amount of substance: by definition, exactly 6.022×10²³, and it is dimensionless. It is named after the scientist Amedeo Avogadro.

It is also known as Avogadro's constant.

∴ Number of particles in one mole = $6.022\times10^2^3$

∴ Number of particles in 2 mole = 2 times Number of particles in one mole

∴ Number of particles in 2 mole= $2\times6.022\times10^2^3=12.044\times10^2^3\approx1.2\times10^2^4$

Hence there are $1.2\times10^2^4$ particles in 2 moles.

4 0

3 years ago

Read 2 more answers