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2 years ago

What is the name of the element with 83 protons and 80 electrons?

1 answer:

8 0

An element with 83 protons has the atomic number 83. If you look at a periodic table you will see Bismuth is the element with the atomic number of 83. It will have a positive charge of +3 because it has only 80 electrons and 83protons. 83-80=+3.

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Mazyrski [523]

Answer:

1 atom of oxygen

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Each water molecule contains 3 atoms making the H2O formula

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2 years ago

Define fundamental unit ? write its two example.

stealth61 [152]

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2 years ago

The automobile beltway around many cities is approximately circular. Suppose that you start driving at point A on the east side

lord [1]

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5 0

2 years ago

Read 2 more answers

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

ivolga24 [154]

Answer: $0.10233nm$

Explanation:

The mean free path $\lambda$ of an atom is given by the following formula:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is the Universal gas constant

$T=0\°C=273.115K$ is the absolute standard temperature

$d$ is the diameter of the helium atom

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ absolute standard pressure

Knowing this, let's find $d$ from (1), in order to find the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius is half the diameter:

$r=\frac{d}{2}$ (5)

Then:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

However, we were asked to find this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Finally:

$r=0.10233nm$ This is the radius of the helium atom in nanometers.

5 0

2 years ago

A 20000 kg subway train initially traveling at 18.5 m/s slows to a stop in a station and then stays there long enough for its br

hodyreva [135]

Answer:

rise the air temperature is 0.179241 K

Explanation:

Given data

mass = 20000 kg

velocity = 18.5 m/s

long = 65 m

wide = 20 m

height = 12 m

density of the air = 1.20 kg/ m³

specific heat = 1020 J/(kg*K)

to find out

how much does the air temperature in the station rise

solution

we know here Energy lost by the train that is calculated by

loss in the kinetic energy that is = 1/2 m v²

loss in the kinetic energy = 0.5 × 20000 ×18.5²

loss in the kinetic energy is 3422500 J

and

this energy is used here to rise the air temperature that is KE / ( specific hat × mass )

so here

air volume = 65 ×20×12

air volume = 15600 m³

air mass = ρ × V = 1.2 × 15600

air mass = 18720 kg

rise the air temperature = 3422500 / ( 1020 × 18720)

rise the air temperature is 0.179241 K

7 0

2 years ago