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3 years ago

The most reactive metals are located at the of the periodic table.

Physics

2 answers:

NeX [460]3 years ago

7 0

Answer:

the elements towards the bottom left corner

Colt1911 [192]3 years ago

4 0

Explanation:

Metals are the substances which tend to lose their valence electrons in order to attain stability.

For example, metals of group 1 are lithium, sodium, potassium, rubidium and cesium.

And, each of these elements have only 1 valence electron present. So, it is much easier to lose just one electron rather than losing 2 or 3 valence electrons.

Hence, alkali metals are the most reactive substances of the periodic table and they are present at extreme left side of periodic table.

Therefore, we can conclude that the most reactive metals are located at the extreme left of the periodic table.

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Given the isotope 2Fes, which has an actual mass of 55.934939 u: a) b) Determine the mass defect of the nucleus in atomic mass u

SSSSS [86.1K]

Answer:

Mass defect of each iron-56 nuclei:

The binding energy per nucleon of Iron-56 is approximately 8.6 MeV.

Explanation:

According to the physics constants table on Chemistry Libretexts:

Proton rest mass: $\rm 1.0072765\;amu$ ;
Neutron rest mass: $\rm 1.0086649\; amu$ .
Speed of light in vacuum: $\rm 2.99792458\times 10^{8}\;m\cdot s^{-1}$ .
Charge on an electron: $\rm 1.6021765\times 10^{-19}\;C$ .

The mass defect of a nucleus is equal to the sum of the mass of its parts (protons and, in most cases, neutrons) minus the mass of the nucleus.

The atomic number of iron is 26. There are 26 protons in each iron-56 nucleus. The mass number 56 indicates that there are 56 nucleons (neutrons and protons) in each iron-56 nucleus. The other $56 - 26 = 30$ particles are neutrons.

The mass of protons and neutrons in each iron-56 nucleus will be:

$\rm 26 \times 1.0072765 + 30 \times 1.0086649 = 56.464736\;amu$ .

According to this question, the mass of an iron-56 nucleus is equal to 55.934939 amu. The mass defect will be

$\rm 56.464736 - 55.934939 = 0.514197\;amu$ .

By the mass-energy equivalence,

$E = m\cdot c^{2}$ .

Refer to this equation, the speed of light in vacuum $c^{2}$ is the conversion factor between mass $m$ and energy $E$ . The value of $c$ is usually given only in SI units $\rm m\cdot s^{-1}$ . Accordingly, the value of $c^{2}$ will be in the SI unit $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ .

Convert million electron-volts to joules.

One electron-volt is equal to the electrical work done moving an electron across a potential difference of one volt.

$\begin{aligned}\rm 1 MeV&= \rm 10^{6}\; eV\\ &= \rm (10^{6}\times 1.6021765\times 10^{-19}\;C)\times 1\; V\\&=\rm 1.6021765\times 10^{-19}\;J\end{aligned}$ .

Convert the unit of $c^{2}$ from $\rm m^{2}\cdot s^{-2} = J\cdot kg^{-1}$ to the desired $\rm MeV \cdot amu^{-1}$ :

$\begin{aligned}c^{2} &= \rm {\left(2.99792458\times 10^{8}\;m\cdot s^{-1}\right)}^{2}\\&=\rm 8.987551787\times 10^{16}\; m^{2}\cdot s^{-2}\\ &= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\\&= \rm 8.987551787\times 10^{16}\; J\cdot kg^{-1}\times \frac{1\;MeV}{1.6021765\times 10^{-13}\;J}\times \frac{1\times 10^{-3}\;kg}{6.022142\times 10^{23}\;amu}\\&\approx \rm 931.602164\;MeV\cdot amu^{-1}\end{aligned}$ .

Total binding energy in each iron-56 nucleus:

$\begin{aligned}E &= m\cdot c^{2}\\&= \rm 0.514197\;amu \times 9.31602164\;MeV\cdot amu^{-1} \\&=\rm 479.027038\; MeV \end{aligned}$ .

Again, the mass number 56 indicates that there are 56 nucleons in each iron-56 nucleus. The binding energy per nucleon of iron-56 $\mathrm{^{56}Fe}$ will be:

$\displaystyle \rm \frac{479.027038\; MeV}{56} \approx 8.6\; MeV$ .

6 0

3 years ago

(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the

frosja888 [35]

Answer:

a.6 V

b.24 V

Explanation:

We are given that

a. $C=8\mu F=8\times 10^{-6} F$

$1\mu =10^{-6}$

Q= $48\mu C=48\times 10^{-6} C$

We know that

$V=\frac{Q}{C}$

Using the formula

$V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V$

b. $Q=192\mu C=192\times 10^{-6} C$

$V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V$

8 0

3 years ago

An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts

Alex Ar [27]

Answer:

y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

F = m a

a = F / m

a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

x= 29 mm = 29 10⁻³ m

On the x axis

v = x / t

t = x / v

t = 29 10⁻³ / 1.7 10⁷

t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

y = $v_{oy}$ t + ½ a t²

y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

y = 77.74 10⁻⁵ m

3 0

3 years ago

Read 2 more answers

What could be done to change this carbide ion to a neutral carbon atom?

Anarel [89]

The answer is to remove 4 electrons.

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3 years ago

How are weather satellites and weather stations similar?

alexandr402 [8]

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6 0

3 years ago