Answer:
<h3>the charge is +1 </h3>
Explanation:
<h3>as we know nutral atom have equal number of protons and electrons</h3><h3>from the give this element have 11 protons so if it is nutral it must have 11 electrons,but in the question this atom is charged this means it gains or losts certain amount of electrons , this atom has 11 proton and 10 electron from this we can understand this atom dicreases by 1 from its proton, this means it losts one electron .</h3><h3>when an atom lostes electron it's charge become positive with the number of electrons it lostes .</h3><h3>this atom lost 1 electron there fore it have +1 charge and become ion called cathion</h3>
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
First, we need to determine the required moles of CaCl₂. We have 500 mL (0.500 L) of a 0.360 M solution (0.360 moles of CaCl₂ per liter of solution).
Then, we will convert 0.180 moles to grams using the molar mass of CaCl₂ (110.98 g/mol).
To prepare the solution, we weigh 20.0 g of CaCl₂ and add it to a beaker with enough distilled water to dissolve it. We stir it, heat it if necessary, and when we have a solution, we transfer it to a 500 mL flask and complete it to the mark with distilled water.
We need to measure 20.0 grams of CaCl₂ to prepare 500 mL of 0.360 M solution.
You can learn more about solutions here: brainly.com/question/2412491
Mass to moles
5.2 mol/Ca(no3)2 to mol
5.2 mol/Ca(no3)2 / molar mass
5.2 mol/Ca(no3)2 / 164.1= 0.032 g/Ca(no3)2
Answer:
1. matches with elements.
2. matches with compounds.
3. matches with atoms
4. matches with weight
5. matches with gas
6. matches with carbon dioxide
7. matches with Mendeleev (there's an element named after him)
8. matches with IUPAC
Hope that helped :)
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
