Molecular mass of Mg(OH)2
= Atomic mass of Mg + 2(Atomic mass of O) + 2(Atomic mass of H)
= 24 g/mol + 2(16 g/mol) + 2(1 g/mol )
= 58 g/mol
Molecular mass of Fe2O3
= 2(Atomic mass of Fe) + 3(Atomic mass of O)
= 2(56 g/mol) + 3(16 g/mol)
= 160 g/mol
Apex the game??? Can o have brainliest
I don't know the options but usually a small strainer or a coffee thing u put over a cup and let the water seep down and the sugar stays.
The given sentence is part of a longer question.
I found this question with the same sentence. So, I will help you using this question:
For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>). If </span>Kp = 0.15, which statement is true of the reaction mixture before
any reaction occurs?
(a) Q = K<span>; The reaction </span>is at equilibrium.
(b) Q < K<span>;
The reaction </span>will proceed to
the right.
(c) Q > K<span>; The reaction </span>will proceed to the left.
The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left,
since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:
Kp is the equilibrium constant in term of the partial pressures of the gases.
Q is the reaction quotient. It is a measure of the progress of a chemical reaction.
The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.
At equilibrium both Kp and Q are equal. Q = Kp
If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,
If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.
Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.
Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
I believe/thought they were very similar due to the fact that they both undergo a process called “oxidation” where they release oxygen into the atmosphere.
