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Luba_88 [7]
2 years ago
15

What liquid is got from copper ii and Ammonia gas?​

Chemistry
1 answer:
11111nata11111 [884]2 years ago
6 0

Answer:

deep blue solution of tetramminecopper [Cu(NH3)4]2+ complex ion.

Explanation:

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A piece of wood with a density of 0.85 g/cm^3 has a mass of 25 g. what is the volume of the wood?
Karolina [17]
Density = m/v  therefore    v = m/d     v = 25/0.85    v = 29.4117... cm^3 ( = 2.94 * 10^-5 m^3
5 0
3 years ago
Read 2 more answers
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
3 years ago
A tank with volume of 2.4 cu ft is filled with Methane to a pressure of 1500 psia at 104 degrees F. Determine the molecular weig
Soloha48 [4]

Explanation:

It is known that equation for ideal gas is as follows.

               PV = nRT

The given data is as follows.

     Pressure, P = 1500 psia,     Temperature, T = 104^{o}F = 104 + 460 = 564 R

     Volume, V = 2.4 cubic ft,      R = 10.73 psia ft^{3}/lb mol R

Also, we know that number of moles is equal to mass divided by molar mass of the gas.

                n = \frac{mass}{\text{molar mass}}

            m = n \times W

                = 0.594 \times 16.04

                = 9.54 lb

Hence, molecular weight of the gas is 9.54 lb.

  • We will calculate the density as follows.

                d = \frac{PM}{RT}

                    = \frac{1500 \times 16.04}{10.73 \times 564}

                    = 3.975 lb/ft^{3}

  • Now, calculate the specific gravity of the gas as follows.

  Specific gravity relative to air = \frac{\text{density of methane}}{\text{density of air}}

                         = \frac{3.975 lb/ft^{3}}{0.0765 lb/ft^{3}}

                         = 51.96

6 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
Calculate the number of chlorine atoms that are present in 66.05 g of dichloromethane, ch2cl2. when you have the number, take it
Whitepunk [10]
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of chlorine = 35.5 grams
Therefore,
molar mass of CH2Cl2 = 12 + 2(1) + 2(35.5) = 85 grams

number of moles = mass / molar mass
number of moles of CH2Cl2 = 66.05 / 85 = 0.777 moles

One mole of CH2Cl2 contains two moles of Cl and each chlorine mole has Avogadro's number of atoms in it.
Therefore,
number of chlorine atoms in 0.777 moles of CH2Cl2 can be calculated as follows:
number of atoms = 0.777 * 2 * 6.022 * 10^23 = 9.358 * 10^23 atoms

Now, we will take log base 10 for this number:
log (9.358 * 10^23) = 23.97119
5 0
3 years ago
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