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4 years ago

PLEASE HELPPP ._. xd

1 answer:

6 0

1. Hyaline: most common; found in the ribs, nose, trachea. Is a precursor of bone
2. Fibro: found in invertebral discs, joint capsules, and ligaments.
3. Elastic: found in the external ear, epiglottis, and larynx.

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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

When the temperature in a room increases from 25°C to 33°C, changes from a solid to a liquid.

mr Goodwill [35]

Answer:

true

Explanation:

Because ice melts if the temperature increasese

8 0

3 years ago

What volume would 3.01•1023 molecules of oxygen gas occupy at STP?

Sliva [168]

First, find moles of oxygen gas: (3.01 x10^23 molec.)/(6.02 x10^23) =0.5mol O2

Second, multiply moles by the standard molar volume of a gas at STP:(0.5mol)(22.4L) = 11.2L O2

6 0

3 years ago

explain why it is a common laboratory procedure to heat analytical reagents and store them in a dessicated atmosphere (a sealed

Readme [11.4K]

Explanation:

Most reagent forms are going to absorb water from the air; they're called "hygroscopic". Water presence can have a drastic impact on the experiment being performed For fact, it increases the reagent's molecular weight, meaning that anything involving a very specific molarity (the amount of molecules in the final solution) will not function properly.

Heating will help to eliminate water, although some chemicals don't react well to heat, so it shouldn't be used for all. A dessicated environment is simply a means to "dry." That allows the reagent with little water in the air to attach with.

6 0

3 years ago

The half life of a certain radioactive element is 800 years. How old is an object if only 12.5% of radioactive atoms in it remai

Levart [38]

Given, half life of a certain radioactive element = 800 years.

Amount of substance remaining at time t = 12.5%

Lets consider the initial amount of the radioactive substance = 100%

Using the half life equation:

A = A₀(1/2)^t/t₁/₂

where A₀ is the amount of radioactive substance at time zero and A is the amount of radioactive substance at time t, and t₁/₂ is the half-life of the radioactive substance.

Plugging the given data into the half life equation we have,

12.5 = 100 . (1/2)^t/800

12.5/100 = (1/2)^t/800

0.125 = (0.5)^t/800

(0.5)^3 = (0.5)^t/800

3 = t/800

t = 2400 years

Thus the object is 2400 years old.

6 0

4 years ago