Answer:
Yes
Explanation:
Masses for the three subatomic particles can be expressed in amu (atomic mass units) or grams. For simplicity, we will use the amu unit for the three subatomics. Both neutrons and protons are assigned as having masses of 1 amu each.
Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:
Now, find moles of 
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
Answer:
Electronegativity is a measure of an atom's ability to attract shared electrons to itself. On the periodic table, electronegativity generally increases as you move from left to right across a period and decreases as you move down a group.
Explanation:
Hope it is helpful....
Answer:
<h3>1. 10 e⁻</h3>
Oxidation numbers
I₂O₅(s): I (5+); O(2-)
CO(g): C(2+); O(2-)
I₂(s): I(0)
CO₂(g): C(4+); O(2-)
<h3>2. 4 e⁻</h3>
Oxidation numbers
Hg²⁺(aq): Hg(2+)
N₂H₄(aq): N(2-); H(1+)
Hg(l): Hg(0)
N₂(g): N(0)
H⁺(aq): H(1+)
<h3>3. 6 e⁻</h3>
Oxidation numbers
H₂S(aq): H(1+); S(2-)
H⁺(aq): H(1+)
NO₃⁻(aq): N(5+); O(2-)
S(s): S(0)
NO(g): N(2+); O(2-)
H₂O(l): H(1+); O(2-)
Explanation:
In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.
1. I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)
Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)
Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻
2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)
Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻
Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)
3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)
Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻
Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O
The number of waves passing in a certain amount of time
