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2 years ago

What happens to a cell placed in an isotonic solution

Chemistry

1 answer:

serg [7]2 years ago

7 0

Answer:

The cell's volume will remain stable.

Explanation:

If a cell is placed in an isotonic solution, there will be no net flow of water into or out of the cell, and the cell's volume will remain stable. If the solute concentration outside the cell is the same as inside the cell, and the solutes cannot cross the membrane, then that solution is isotonic to the cell.

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People mine zinc metal for use as an anticorrosive, skin protectant, and in electrical components. Zinc mining involves using ch

ankoles [38]

C I’m pretty sure. I hope this helps

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3 years ago

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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.

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4 years ago

1) Which of the following is NOT a step in the scientific method *

Alex_Xolod [135]

Answer:

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3 years ago

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An isotope of iron has 28 neutrons. If the atomic mass of the isotope is 54, how many protons does it have?

Crazy boy [7]

26 because you need to do 54-28=26
The atomic mass number is the number of protons AND neutrons together.

5 0

3 years ago

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Which pair has identical electron configurations?

Thepotemich [5.8K]

Identical electron configurations : K⁺ and Cl⁻

<h3>Further explanation </h3>

In an atom, there are levels of energy in the shell and sub-shell

This energy level is expressed in the form of electron configurations.

Charging electrons in the sub-shell uses the following sequence:

S²⁻ and CI

S²⁻ : [Ne] 3s²3p⁶

Cl : [Ne] 3s²3p⁵

K⁺ and CI⁻

K⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶

Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶

S and Ar

S :[Ne] 3s²3p⁴

Ar : [Ne] 3s²3p⁶

Cl⁻ and K

Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶

K : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹

6 0

3 years ago