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lapo4ka [179]
3 years ago
5

In what situations are van der waals forces the most significant

Chemistry
1 answer:
viktelen [127]3 years ago
7 0
Its most significant when all other forces are absent
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Theoretically, what mass of [Co(NH3)4(H2O)2]Cl2 could be produced from 4.00 g of CoCl2•6H2O starting material. If 1.20 g of [Co(
Romashka [77]

<u>Answer:</u> The theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CoCl_2.6H_2O = 4.00 g

Molar mass of CoCl_2.6H_2O = 238 g/mol

Putting values in equation 1, we get:

\text{Moles of }CoCl_2.6H_2O=\frac{4.00g}{238g/mol}=0.0168mol

The chemical equation for the reaction of CoCl_2.6H_2O to form  [Co(NH_3)_4(H_2O)_2]Cl_2 follows:

CoCl_2.6H_2O+4NH_3\rightarrow [Co(NH_3)_4(H_2O)_2]Cl_2+4H_2O

By Stoichiometry of the reaction:

1 mole of CoCl_2.6H_2O produces 1 mole of [Co(NH_3)_4(H_2O)_2]Cl_2

So, 0.0168 moles of CoCl_2.6H_2O will produce = \frac{1}{1}\times 0.0168=0.0168mol of [Co(NH_3)_4(H_2O)_2]Cl_2

Now, calculating the mass of [Co(NH_3)_4(H_2O)_2]Cl_2 from equation 1, we get:

Molar mass of [Co(NH_3)_4(H_2O)_2]Cl_2 = 234 g/mol

Moles of [Co(NH_3)_4(H_2O)_2]Cl_2 = 0.0168 moles

Putting values in equation 1, we get:

0.0168mol=\frac{\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2}{234g/mol}\\\\\text{Mass of }[Co(NH_3)_4(H_2O)_2]Cl_2=(0.0168mol\times 234g/mol)=3.93g

To calculate the percentage yield of [Co(NH_3)_4(H_2O)_2]Cl_2, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 1.20 g

Theoretical yield of [Co(NH_3)_4(H_2O)_2]Cl_2 = 3.93 g

Putting values in above equation, we get:

\%\text{ yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=\frac{1.20g}{3.93g}\times 100\\\\\% \text{yield of }[Co(NH_3)_4(H_2O)_2]Cl_2=30.53\%

Hence, the theoretical yield and percent yield of [Co(NH_3)_4(H_2O)_2]Cl_2 is 3.93 g and 30.53 % respectively

6 0
2 years ago
What volume of O2(
irakobra [83]
Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
<span>R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.</span>
V(O₂) = n·R·T / p.<span>
V(O₂) = 0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.</span><span>
V(O₂<span>) = 4.63 L.</span></span>
4 0
3 years ago
D‑Glucose and L‑glucose are
tensa zangetsu [6.8K]

Answer:

<h3>A. Epimers </h3>

Explanation:

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3 years ago
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I need help ASAP <br> Extra points &amp; brainlest:)
Oksana_A [137]

Answer:

it is c

Explanation:

3 0
2 years ago
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A tank of gas is found to exert 8.6 atm at 38°C. What would be the required
Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 8.6 atm
  • Initial temperature of the gas (T₁): 38°C
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0
2 years ago
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