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3 years ago

How could you demonstrate boiling point?

1 answer:

5 0

A light layer of vacuum grease is applied to the rim of the belljar. Water at room temperature is placed inside and the vacuum pump is then used to evacuate the vessel. When the air pressure is reduced to the vapour pressure of water at room temperature the water will begin to boil.

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Which type of chemical bond involves the exchange of electrons

Alex17521 [72]

Answer:

Ionic or Electrovalent Bonding

Explanation:

There are primarily two categories of bonding between chemical entities. We have; Ionic Bonding and Covalent Bonding.

Ionic bonding or electrovalent bonding is the complete transfer of valence electron(s) between atoms. There is the transfer of electron from typically a metal to a non metal.

Covalent Bonding however involves the sharing of electrons between atoms. Depending on whuch atoms provide the electrons, it can be ordinary covalent oor coordinate covelent bond.

5 0

3 years ago

How many moles of helium are needed to fill a balloon to a volume of 5.3 L at 22 ℃ and 632 mmHg?

son4ous [18]

Answer:

0.18 moles

Explanation:

Applying,

PV = nRT................... Equation 1

Where P = pressure, V = volume, n = number of moles, R = molar gas constant, T = temperature.

make n the subject of the equation

n = PV/RT............... Equation 2

Given: V = 5.3 L, T = 22 °C = (22+272) K = 295 K, P = 632 mmHg = (0.00131579×632) = 0.8316 atm, R = 0.083 L.atm/K.mol

Substitute these values into equation 2

n = (0.8316×5.3)/(0.083×295)

n = 0.18 moles

6 0

3 years ago

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

Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0

3 years ago

3. What determines the mass number of an element? The mumhor of nrotons

ArbitrLikvidat [17]

Answer: The number of protons and the number of neutrons determine an element's mass number.

Explanation:

5 0

3 years ago

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