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4 years ago

A solution has a ph of 10.20 at 25°c. what is the hydroxide-ion concentration at 25°c?

Chemistry

1 answer:

kenny6666 [7]4 years ago

5 0

Using equation:

pH+pOH=14

We are given pH=10.20 at 25°c.

So putting this value in the equation,

We have pOH=14-10.20

= 3.8

pOH=-log(OH)

-log(OH)=3.8

OH=0.00015M

So hydroxide concentration will be 0.00015 M.

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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

1) What is the electron configuration of the element with 27 protons?

Romashka [77]

1) cobalt u can tell bc on a periodic table there is a small number that on cobalt is 27.

2) think that would be 11 bc in the 4th shell there can be up to 18 electrons

6 0

3 years ago

A cube of an unknown metal is 1.25 inches on each side and has a mass of 128.3741 g. What is its' density?

Mademuasel [1]

Hey there!

To find the density of an object, you must use this formula:
Density=Mass/Volume

Knowing that your mass is 128.3741, the only information you need left would be to find the volume of the cube.

Because the side length of the cube is given, you can multiply the length three times in order to find its volume:
1.25*1.25*1.25
=1.953125

Now that you have your volume and mass, divide the mass by the volume to find the density:
128.3741/1.953125
=65.7275392

Therefore, your density would be 65.7275392 grams per inches cubed.

6 0

3 years ago

calculate the molality of commercial HCl solution which is 12.1Molarity, and has a density of 1.19g/mL, and is 37.2wt.%HCl.

shutvik [7]

Answer:

0.44165 Molal

8 0

4 years ago

You may expect _____

TiliK225 [7]

Answer:

covalent..............

5 0

3 years ago