If 28.0 grams of a gas occupies 22.4 liters at STP, the gas could be carbon monoxide, CO
<h3>Ideal gas </h3>
We understood from the ideal gas equation that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP)
<h3>How to determine the identity of the gas</h3>
To determine the identity of the gas, we shall determine the mass of 1 mole of each gas. This can be obtained as
For C₂H₂
1 mole of C₂H₂ = (12×2) + (2×1) = 26 g
For C₂H₆
1 mole of C₂H₆ = (12×2) + (6×1) = 30 g
For CO₂
1 mole of CO₂ = 12 + (16×2) = 44 g
For CO
I mole of CO = 12 + 16 = 28 g
From the above illustrations, we can see that 1 mole of CO is equivalent to 28 g.
Thus, the correct answer to the question is CO
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it should be True unless if I’m wrong
Answer:
The length and strength of the C-H bond are different for ethine, ethylene and ethane due to the difference between the s characters (sp 3, sp 2, sp) in each chemical structure, which gives stronger bonds. For example, the length of the C-H bond in ethino is 1.06Å, which is also less than in ethane (1.09Å) or ethylene (1.08Å).
Explanation:
Answer: 1 × 10−2 M
Explanation: Kw is known as the constant of water dissociation and its value is found to be 
. It is the product of the concentration of [H3O+] and [OH -] ions .
Dissociation of water can be represented by -
Kw = [H3O+] [OH -]
As we know that pKw = - log Kw,
Thus pKw = 14
pOH = - log OH- = - log (10^{-12}) = 12
and pKw = pH + pOH
14 = pH + 12
pH = 14 - 12 = 2
pH = -log H3O+
2 = -log H3O+
Therefore, [H3O+] = 1 × 10−2 M
