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3 years ago

30 points

2 answers:

8 0

The molecular weight for Calcium Chloride Dihydrate (CaCl2•2H20) is 2H2O.

mass of 64g/mol = 256.52 g/mol
mass of 11g/mol = 26.981538 g/mol
mass of 147g/mol= 146.9149 g/mol
258g/mol = 258.09843 g/mol

Tcecarenko [31]3 years ago

4 0

Answer:

$M=147.014 g/mol$

Explanation:

The atomic weights of the elements:

Ca=40.078 g/mol

Cl=35.453 g/mol

H=1.008 g/mol

O=15.999 g/mol

The calcium chloride dihydrate:

$M=(40.078+35.453*2+2*2*1.008+2*15.999)g/mol$

$M=147.014 g/mol$

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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

3 years ago

A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0- L vessel at 300 K . The following equilibr

umka2103 [35]

Answer:

The concentration of N2 at the equilibrium will be 0.019 M

Explanation:

Step 1: Data given

Number of moles of NO = 0.10 mol

Number of moles of H2 = 0.050 mol

Number of moles of H2O = 0.10 mol

Volume = 1.0 L

Temperature = 300K

At equilibrium [NO]=0.062M

Step 2: The balanced equation

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

Step 3: Calculate the initial concentration

Concentration = Moles / volume

[NO] = 0.10 mol / 1L = 0.10 M

[H2] = 0.050 mol / 1L = 0.050 M

[H2O] = 0.10 mol / 1L = 0.10 M

[N2] = 0 M

Step 4: Calculate the concentration at the equilibrium

[NO] at the equilibrium is 0.062 M

This means there reacted 0.038 mol (0.038M) of NO

For 2 moles NO we need 2 moles of H2 to produce 1 mol N2 and 2 moles of H2O

This means there will also react 0.038 mol of H2

The concentration at the equilibrium is 0.050 - 0.038 = 0.012 M

There will be porduced 0.038 moles of H2O, this means the final concentration pf H2O at the equilibrium is 0.100 + 0.038 = 0.138 M

There will be produced 0.038/2 = 0.019 moles of N2

The concentration of N2 at the equilibrium will be 0.019 M

5 0

3 years ago

Student Exploration: Limiting Reactants Vocabulary: chemical equation, chemical formula, chemical reaction, coefficient, limitin

SIZIF [17.4K]

Answer:

30 hot dogs

Explanation:

It is given that :

There are 4 packets of eight wieners, i.e. 4 x 8 = 32 wieners

There are 3 bags of ten buns, i.e. 3 x 10 = 30 buns

One hot dogs need 1 bun and 1 wiener to make a hot dog.

There are 30 buns, so 30 hot dogs can be made out by using all the 30 buns and the 30 wieners out of the 32 wieners.

Therefore, 30 hot dogs.

And the number of extra wieners left = 32 - 30 = 2 wieners.

6 0

2 years ago

Balance the equation :

Sati [7]

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

<u>Explanation:</u>

Aluminum has a typical oxidation condition of 3+ , and that of iodine is 1- . Along these lines, three iodides can bond with one aluminum. You get AlI3. For comparable reasons, aluminum chloride is AlCl3.

Chlorine and iodine both exist normally as diatomic components, so they are Cl2( g ) also, I2( g ), individually. In spite of the fact that I would anticipate that iodine should be a strong.

Balancing the equation, we get:

2AlI 3( aq ) + 3Cl2 ( g ) → 2AlCl3 ( aq ) + 3 I 2 ( g )

Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the AlCl 3 on the right.

Normally, presently we have two Al on the right, so I multiplied the AlI 3 on the left. Hence, I have 6 I on the left, and I needed to significantly increase I 2 on the right.

We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.

3 0

3 years ago

La formula estructural del eter butil pentilico

Mama L [17]

Answer:

do you mean?

Explanation:

the structural formula of butyl pentyl ether

3 0

3 years ago