All categories

3 years ago

According to atomic theory:

Chemistry

1 answer:

JulsSmile [24]3 years ago

7 0

Answer:

e. all of these

Explanation:

Let us check all the given options one by one:

a.The nucleus is positively charged.

Yes it is correct since nucleus contain protons and neutrons and protons are positively charged.

b.The nucleus contains both charged and uncharged particles.

Yes because protons are positively charged and neutrons are neutral in nature.

c.The electrons contribute very little to the total mass of the atom.

Yes we know all the mass of the atom is considered in center and mass of electron is negligible as compared to protons and neutrons.

d.The electrons are located in the atomic space outside the nucleus.

Yes, its a known fact.

e. All of these.

Since , all given options are correct .

Therefore , option e. is right .

You might be interested in

Is helium a homogeneous or heterogeneous mixture or compound or element?

sattari [20]

Helium is a chemical ELEMENT of the family of noble gases. Its chemical symbol is 'He' which has an atomic number of 2. It is a colorless, odorless, tasteless, non-toxic, inert, monatomic gas, the first in the noble gas group in the periodic table. Although it doesn't really react under normal conditions, it is the second most abundant element in the universe.

Elements are pure substances that cannot be broken down any further.

8 0

3 years ago

Which voice part sings the lowest notes in an SATB choir?

Talja [164]

Hello

Question: which voice sings lowest in a SATB choir?

Answer: A. Bass

Hope this helps and have a great day!

4 0

3 years ago

Which elements usually lose their valence electrons when they bond

masya89 [10]

Metals, for example Cesium and Francium

5 0

3 years ago

Read 2 more answers

What temperature would 3.54 moles of xenon gas need to reach to exert a pressure of 1.57 atm at a volume of 34.6 l

ira [324]

Answer:

186.9Kelvin

Explanation:

The ideal gas law equation is PV = n R T

where

P is the pressure of the gas

V is the volume it occupies

n is the number of moles of gas present in the sample

R is the universal gas constant, equal to 0.0821 atm L /mol K

T is the absolute temperature of the gas

Ensure units of the volume, pressure, and temperature of the gas correspond to R ( the universal gas constant, equal to 0.0821 atm L /mol K )

n = 3.54moles

P= 1.57

V= 34.6

T=?

PV = n R T

PV/nR = T

1.57 x 34.6/3.54 x 0.0821

54.322/0.290634= 186.908620464= T

186.9Kelvin ( approximately to 1 decimal place)

5 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