Answer:
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Answer:
AgNO₃(aq) + NaCl(aq) ⇄ AgCl(s)↓ + NaNO₃(aq) Kps
2K₃PO₄(aq) + 3MgCl₂(aq) ⇄ 6KCl (aq) + Mg₃(PO₄)₂ (s)↓ Kps
Explanation:
In the first reaction (1) we have:
Silver nitrate and sodium chloride as the reactants. Then the products will be, silver chloride (s) and sodium nitrate.
We know that all the salts from the nitrate are soluble, so the chloride will react to the silver cation to make the precipitate, an insoluble salt.
AgNO₃(aq) + NaCl(aq) ⇄ AgCl(s)↓ + NaNO₃(aq) Kps
For the second reaction (2), the reactants are:
Potassium phosphate and magnessium chloride
Phosphate salts are insoluble, except for the group 1. The same as chloride that makes soluble salts with the elements from the group 1.
The equation that makes the precipitate is.
2K₃PO₄(aq) + 3MgCl₂(aq) ⇄ 6KCl (aq) + Mg₃(PO₄)₂ (s)↓ Kps
Answer:
C
Explanation:
the rest are trends except for Electronegativity
<em>Answer:</em>
Symbol = I
<em>Explanation:</em>
<em>Isotopes</em> are those substances which have same atomic number (Numbers of protons) and different masses (Number of proton + numbers of neutrons).
- Iodine also have many isotopes like
<em>Iodine-125</em> that contain 53 protons, 72 neutrons.
<em>Iodine-128</em> that contain 53 protons, 75 neutrons.
<em>Iodine-129 </em> that contain 53 protons, 76 neutrons.
<em>Iodine-131</em> that contain 53 protons, 78 neutrons.
<em>Summary:</em>
- Here the Iodine-131 is correct answer of this question.
In order to determine the number of protons in 20.02 mol of Ne, we use Avogadro's number to convert the number of moles to number of atoms, 1 mol = 6.022 x 10^23 atoms. From there, we must know the number of protons in a Neon atom, which is 10. Thus, the formula will be:
(20.02 mol Ne)x(6.022 x 10^23 atoms/mol)x(10 protons/1 atom Ne) =
1.2056 x 10^26 protons