What is the molarity of Li in a solution that contains of 46.552ppm lithium ferrocyanide Li3fe(cn)6

Calculate the concentration of the following solution in mol/dm3 0.1 moles of NaCl in 200 cm3

taurus [48]

1 $cm ^{3}$ = 0.001 $dm ^{3}$ . Therefore 200 $cm ^{3}$ = 0.2 $dm ^{3}$ . Molarity = $\frac{number of moles of NaCl}{volume of the solution} = \frac{0.1}{0.2} = 0.5 mol/dm^{3}$ . Hope this helps.

8 0

3 years ago

Draw the best Lewis structure for NH3 by filling in the bonds, lone pairs, and formal charges. (Assign bonds, lone pairs, radica

kiruha [24]

Answer : The Lewis-dot structure of $NH_3$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $NH_3$

As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in $NH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on N}=5-2-\frac{6}{2}=0$

$\text{Formal charge on }H_1=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_2=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_3=1-0-\frac{2}{2}=0$

Hence, the Lewis-dot structure of $NH_3$ is shown below.

3 0

3 years ago

In NMR if a chemical shift(δ) is 211.5 ppm from the tetramethylsilane (TMS) standard and the spectrometer frequency is 556 MHz,

Vika [28.1K]

Answer:

The answer is: 11759 Hz

Explanation:

Given: Chemical shift: δ = 211.5 ppm, Spectrometer frequency = 556 MHz = 556 × 10⁶ Hz

In NMR spectroscopy, the chemical shift (δ), expressed in ppm, of a given nucleus is given by the equation:

$\delta (ppm) = \frac{Observed\,frequency (Hz)}{Frequency\,\, of\,\,the\,Spectrometer (MHz)} \times 10^{6}$

$\therefore Observed\,frequency (Hz)= \frac{\delta (ppm)\times Frequency\,\, of\,\,the\,Spectrometer (MHz)}{10^{6}}$

$Observed\,frequency= \frac{211.5 ppm \times 556 \times 10^{6} Hz}{10^{6}} = 11759 Hz$

<u>Therefore, the signal is at 11759 Hz from the TMS.</u>

6 0

3 years ago

ODIO POD

miss Akunina [59]

Answer:

$\boxed{\text{40 mol Al}}$

Explanation:

Al₂O₃ ⟶ 2Al + 3O₂

n/mol: 20

$\text{Moles of Al} = \text{20 mol Al$_{2}$O$_{3}$}\times \dfrac{\text{2 mol Al}}{\text{1 mol Al$_{2}$O$_{3}$}}= \textbf{40 mol Al}\\\text{You can produce }\boxed{\textbf{40 mol Al}}$

8 0

3 years ago

The Synthesis Reaction of aluminum (Al) and iodine (I) forms a new, more complex

marysya [2.9K]

Answer:

Aluminum iodide (AlI₃)

Explanation:

The synthesis reaction of aluminum (Al) and iodine (I) can be illustrated as shown below:

Aluminium exhibit trivalent positive ion (Al³⁺)

Iodine exhibit univalent negative ion (I¯)

During reaction, there will be an exchange of ion as shown below:

Al³⁺ + I¯ —> AlI₃

Thus, we can write the balanced equation for the reaction as follow:

Al + I₂ —› AlI₃

There are 2 atoms of I on the left side and 3 atoms on the right side. It can be balance by putting 2 in front of AlI₃ and 3 in front of I₂ as shown below:

Al + 3I₂ —› 2AlI₃

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by putting 2 in front of Al as shown below:

2Al + 3I₂ —› 2AlI₃

Thus the equation is balanced.

The product on the reaction is aluminum iodide (AlI₃)

3 0

3 years ago