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dimulka [17.4K]
3 years ago
7

2C₂H6 + 702 —>4C02 + 6H₂O

Chemistry
1 answer:
navik [9.2K]3 years ago
8 0

Answer:

975.56×10²³ molecules

Explanation:

Given data:

Number of molecules of C₂H₆ = 4.88×10²⁵

Number of molecules of CO₂ produced  =  ?

Solution:

Chemical equation:

2C₂H₆  + 7O₂     →      4CO₂ + 6H₂O

Number of moles of C₂H₆:

1 mole = 6.022×10²³ molecules

4.88×10²⁵  molecules×1mol/6.022×10²³ molecules

0.81×10² mol

81 mol

Now we will compare the moles of C₂H₆ with CO₂.

                     C₂H₆          :             CO₂

                          2           :               4

                           81         :           4/2×81 = 162 mol

Number of molecules of CO₂:

1 mole = 6.022×10²³ molecules

162 mol ×6.022×10²³ molecules / 1 mol

975.56×10²³ molecules

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A sealed container at 25oC contains a gas at a pressure of 104 kPa. What is the pressure of the gas when it is heated to 225oC?
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174 kPa

Explanation:

Given that,

Initial temperature, T₁ = 25° C = 25+273 = 298 K

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We need to find the new pressure. The relation between the temperature and pressure is given by :

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Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
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Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

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165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

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Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

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Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

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105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

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