Because they both have to do with and chemistry science
Answer is: 5,75·10⁻¹.
Kf = 2,3·10⁶ 1/s.
K = 4,0·10⁸ 1/s.
Kr = ?
Kf - <span>forward rate constant.
K - </span><span>equilibrium constant.
Kr - </span><span>reverse rate constant.
</span>Since both Kf and Kr are constants at a given temperature, their ratio is also a constant that
is equal to the equilibrium constant K.<span>
K = Kf/Kr.
Kr = Kf/K = </span>2,3·10⁶ 1/s ÷ 4,0·10⁸ 1/s = 5,75·10⁻¹.
Explanation:
Ions form when atoms gain or lose electrons. This is so that they form a full outer shell of electrons. When an atom gains electrons it becomes a negative ion, because electrons are negatively charged. For example, all halogens (group 7 or 17) form negative ions as they gain an electron forming a 1- charge. When an atom loses electrons it becomes a positive ion, as it is losing some negative charge from the electrons. This would be for example, alkali metals (group 1) which lose an electron to form a positive ion with a 1+ charge, (ALL metals form positive ions).
The reaction between copper II chloride and sodium sulfide as well as lead II nitrate and potassium sulfate both produce precipitates.
The solubility of a substance in water is in accordance with the solubility rules. It is possible that a solid product may be formed when two aqueous solutions are mixed together. That solid product is referred to as a precipitate.
Now, we will consider each reaction individually to decode whether or not a precipitate is possible.
- In the first reaction, we have; CuCl2(aq) + Na2S(aq) ---->CuS(s) + 2NaCl(aq). A precipitate (CuS) is formed.
- In the second reaction, Pb(NO3)2(aq) + 2KNO3(aq) ----> PbSO4(s) + KNO3(aq), a precipitate PbSO4 is formed
- In the third reaction, NH4Br(aq) + NaOH(aq) ----->NH3(g) + NaBr(aq) + H2O(l), a precipitate is not formed here.
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