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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Answer:
The correct option is: A. H₂SO₄
Explanation:
Acid is a charged or a neutral molecule that is a proton donor and electron pair acceptors.
Acids can be classified into monoprotic acids and polyprotic acids.
Monoprotic acids are the acids that can release only one proton on dissociation.
Whereas, polyprotic acids are the acids that can release more than one proton on dissociation.
Diprotic acid is a type of polyprotic acid that can release two protons on dissociation. Example: H₂SO₄
Answer:
Choice B, C, and D.
Explanation:
Choice A is not true in general. Here's a way to think about that. Consider a very special equilibrium where the concentration of reactants and products are indeed equal. When one of the external factors (such as temperature) changes, the equilibrium will shift towards either side of the reaction. More products will be converted to reactants, or vice versa. Either way, in the new equilibrium, the concentration of the reactants and products will not be equal any more.
Choice B should be considered with choice C and D in mind.
Choice C is indeed correct. The reaction rate would not be zero unless all the reactants were used up or taken out of the system. That's not what happens in an equilibrium. Instead, when reaction rate is plotted against time, the graph for reactions in both directions will eventually flat out at a non-zero value.
Choice D explains why even though choice C is correct, the concentration of a system at equilibrium stays the same. At the equilibrium, reactions in both directions are still happening. However, during the time it takes for the forward reaction use up some reactant particles, the reverse reaction would have produced these particles again. On a large scale, there would be no observable change to the concentration of each species in the equilibrium. Therefore, choice B is also correct.
It is represented by C) KI
K+
I-
So it's KI
Carbon has 6 protons and 8 neutrons.
