Answer:
2NaOH (aq) + CaCl2 (aq) -> 2NaCl(aq) + Ca(OH)2(s)
Formula of precipitate: Ca(OH)2 <em>(s)</em>
Explanation:
First, we do the double replacement reaction to determine our chemical equation between the reactants and products. Once we have our products, with a solubility chart (I added one below) we can determine which of the products is soluble or insoluble.
In this case NaCl is soluble or aqueous (meaning it can dissolve in water) and Ca(OH)2 is insoluble (meaning that when the reactions takes place, these two will form a solid/precipitate)
<span>You can answer this question by getting the atomic number and atomic mass of Oxygen from a periodic table. There you will find that the atomic number is 8, that means, by definition, that it has 8 protons. This is, because atomic number is defined as the number of protons of an element. Given that the atom is neutral, that implies that the atoms have the same number of electrons than protons. So you already know that the oxygen atoms has 8 protons and 8 electrons. The number of neutrons can vary, which is what defines the isotopes. Given that the atomic mass of oxygen is 15.999, that means that most atoms of oxygen has 8 neutrons (8 protons +8 neutrons = 16 atomic mass). But you can not be sure that a specific atom of oxygen has 8 neutrons, nevertheless, given that the other options are discarded (because they do not have 8 protons and 8 electrons), the only correct answer is the option A. 8 protons, 8 electrons, and 8 neutrons.</span>
Answer:
ⁿₐX => ²¹⁸₈₄Po
Explanation:
Let ⁿₐX be the isotope.
Thus, the equation can be written as follow:
²²²₈₆Rn —> ⁴₂α + ⁿₐX
Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:
222 = 4 + n
Collect like terms
222 – 4 = n
218 = n
Thus,
n = 218
86 = 2 + a
Collect like terms
86 – 2 = a
84 = a
Thus,
a = 84
ⁿₐX => ²¹⁸₈₄Po
²²²₈₆Rn —> ⁴₂α + ⁿₐX
²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po
Answer:
The answer to your question is: 69.6 %
Explanation:
Freon -112 (C₂Cl₄F₂)
MW = (12 x 2) + (35.5 x 4) + (19 x 2)
= 24 + 142 + 38
= 204 g
204 g of C₂Cl₄F₂ ----------------- 100%
142 g ----------------- x
x = (142 x 100 ) / 204
x = 69.6 %
