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3 years ago

How many aluminum atoms are in 2.58 grams of aluminum

Chemistry

1 answer:

Ket [755]3 years ago

6 0

Answer:

5.75x10^22 atoms

Explanation:

From a clear understanding of Avogadro's hypothesis, we discovered that 1mole of any substance contains 6.02x10^23 atoms. This implies that 1mole of Al also contains 6.02x10^23 atoms.

1mole of Al = 27g

If 27g of Al contains 6.02x10^23 atoms, then 2.58g of Al will contain = (2.58 x 6.02x10^23)/27 = 5.75x10^22 atoms

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Luke and Sian want to plant a vegetable garden in their yard. A soil testing kit measures the soil pH at 5.0, but the lettuce th

dexar [7]

Answer:

So, Luke and Sian has to increase the pH of the soil by adding base to it.

Explanation:

The pH is defined as the negative logarithm of the hydrogen ion concentration in their aqueous solution.

$pH=-\log[H^+]$

With increase in hydrogen ion concentration the pH value decreases.
With decrease in hydrogen ion concentration the pH value increases.

The pH of the soil after testing it on a kit comes out be 5.0, but they both need pH of the soil to 6.5.

Comparison of pH of soil:

= 5.0 < 6.5

= High hydrogen ion concentration > High hydrogen ion concentration

So, Luke and Sian has to increase the pH of the soil by adding base .Doing so will decrease the hydrogen ion concentration in the soil (where as addition of acid lower the pH of soil).

4 0

3 years ago

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

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Dovator [93]

Answer: 406 hours

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000$ Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = $\frac{193000}{63.5}\times 19000=57748032$ Coloumb

$57748032=39.5\times t$

$t=1461975sec=406hours$ (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

3 0

3 years ago

The smith family uses more energy in the summer than the winter what is a good explanation for this?

DENIUS [597]

They use fans and air conditioning in the summer because its hot, which means they use energy and in the winter, there is no need because its cold so less energy.

6 0

3 years ago

Read 2 more answers

An abandoned Indiana coal mine spoil bank (wastes) contains chunks of pyrite minerals. Under constant erosion and weathering, th

Minchanka [31]

Answer:

The acid will be neutralized overtime

Explanation:

The presence of the pyrites leads to the leaching of large amounts of sulphuric acid, however the basic carbonates neutralizes the acid according to the reaction equation;

CaCO3 + H2SO4 ---> CaSO4 + CO2 + H2O.

This will prevent all the deleterious consequences associated with the leaching of the acid in the abandoned coal mine.

7 0

3 years ago