Answer:
Explanation:
The question is asking to write the <em>isotopic symbol </em>of the form 
for the <em>sodium isotope with 13 neutrons</em>.
In general, the isotopic symbol in the given form shows:
- The element's chemical symbol: X
- A: mass number of the isotope, written as a superscript to the left of the element's simbol, and
A = number of protons + number of neutrons
- Z: atomic number of the isotope, written as a subscript to the left of the elements's symbol,
Z = number of protons
The atomic symbol of sodium is Na.
The atomic number, or number of protons, is the same for every isotope of the element, and you can find it in any periodic table. Tha atomic number of sodium is 11. Thus:
The mass number is the number of protons plus neutrons, hence:
Now you can write the isotope symbol for the sodium isotope with 13 neutrons:
Ir, Iridium has 77 electrons when neutral.
I need points lol sorry for wasting ur time we
To figure out questions related to reacting moles/masses, the first step is always to write a complete balanced equation.
2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)
Since Cl2 is the excess reactant, and Fe is the limiting reactant, we can simply find the number of moles of the product by comparing the mole ratio of the limiting reactant to the mole ratio of the product from the equation.
From the equation, mole ratio of Fe:FeCl3 = 2:2 = 1:1, the number of moles of product is exactly the same as the number of moles of the limiting reactant, which makes it 8 moles.
Note that if the mole ratio is not 1:1, you have to do some calculations to make sure the no. of moles is balanced at the end. Which means, if the mole ratio happened to be 1:2, the no. of moles of the product would be 8x2=16 instead.
So, your answer is 8 moles.
Answer:
5.83 mol.
Explanation:
- From the balanced reaction:
<em>2Al + 3Ag₂S → 6Ag + Al₂S₃,</em>
It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.
Al reacts with Ag₂S with (2: 3) molar ratio.
<em>So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.</em>
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- The limiting reactant is Ag₂S.
- The excess "left over" reactant is Al.
The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.
