The answer is: 3.61·10²⁴ atoms of oxygen.
n(NaNO₃) = 2.00 mol; amount of sodium nitrate.
In one molecule of sodium nitrate, there are three atoms of oxygen.
n(O) = 3 · 2 mol.
n(O) = 6.00 mol; amount of oxygen atoms.
N(O) = n(O) · Na.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(O) = 6 mol · 6.022·10²³ 1/mol.
N(O) = 3.61·10²⁴; number of oxygen atoms.
Answer:
The answer is
<h2>686.01 g</h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
volume of zinc = 96.08 cm³
density = 7.14 g/cm³
The mass is
mass = 96.08 × 7.14 = 686.0112
We have the final answer as
<h3>686.01 g</h3>
Hope this helps you
Answer:
Explanation:
Each magnesium atom loses two electrons and each iodine atom gains one electron. So there should be a 1-to-2 ratio of magnesium ions to iodide ions.
Answer:
Right
Explanation:
The given parameters are;
The soluble sulfate formed by the metal, M = M₂SO₄
In the galvanic cell formed by the metal M, we have;
The concentration of M₂SO₄ in the left half cell = 50.0 mM = 0.05 M
The concentration of M₂SO₄ in the right half cell = 5.00 M
In the galvanic cell, the metal 'M' will be dissolved into the solution with lower concentration as M²⁺ which is the left half cell, making the cell negative and the solution more concentrated
In the right half cell, the metal 'M²⁺' in the solution will be plated unto the electrode making the solution less concentrated and the electrode in the right half cell will be the positive electrode
Therefore;
The electrode which will be positive is the electrode in the right half cell.
The balanced equation for the reaction is as follows
2Na + Cl₂ --> 2NaCl
first we need to find the limiting reactant from the 2 reactants in the reaction
stoichiometry of Na to Cl₂ is 2:1
number of Na moles - 133.0 g / 23 g/mol = 5.783 mol
at STP, molar volume is where 1 mol of any gas occupies a volume of 22.4 L
if 22.4 L occupied by 1 mol
then 33.0 L is occupied by 33.0 L / 22.4 L/mol = 1.47 mol
therefore number of Cl₂ moles present - 1.47 mol
If Cl₂ is the limiting reactant
1 mol of Cl₂ reacts with 2 mol of NaCl
therefore 1.47 mol of Cl₂ reacts with - 2 x 1.47 = 2.94 mol of Na
but 5.783 mol of Na is present which means Na is in excess and Cl₂ is the limiting reactant
NaCl formed depends on amount of Cl₂ present
stoichiometry of Cl₂ to NaCl is 1:2
number of NaCl moles formed - 2 x 1.47 mol = 2.94 mol of NaCl
mass of NaCl formed - 2.94 mol x 58.5 g/mol = 172 g
172 g of NaCl is formed