Answer:
The partial pressure of argon in the flask = 71.326 K pa
Explanation:
Volume off the flask = 0.001 
Mass of the gas = 1.15 gm = 0.00115 kg
Temperature = 25 ° c = 298 K
Gas constant for Argon R = 208.13 
From ideal gas equation P V = m RT
⇒ P = 
Put all the values in above formula we get
⇒ P = 
× 208.13 × 298
⇒ P = 71.326 K pa
Therefore, the partial pressure of argon in the flask = 71.326 K pa
Tons of wind, rain, thunder. Stuff like that.
Explanation:
The given reaction is redox reaction but not combustion reaction.
Redox reaction : It is a chemical reaction in which oxidation and reduction occurs simultaneously.Oxidation is gaining of electrons and reduction is loosing of electrons.
oxidation
reduction
Potassium when reacts with water ,gives potassium ion in aqueous medium and combines with hydroxide ion present in water to form potassium hydroxide where as 
ions of water by gaining electrons (from potassium in water) gives hydrogen gas.
Hence , yes it is a redox reaction
Combustion reaction are the chemical reaction in which one reactant react with oxygen molecule to give heat and light.
Here in given reaction no oxygen molecule is appearing on reactant side.
Hence not a combustion reaction.
Answer:
0.726 mol·L⁻¹
Step-by-step explanation:
c = moles/litres
=====
Moles = 29.8 × 1/342.30
Moles = 0.087 06 mol
=====
Litres = 120 × 1/1000
Litres = 0.120 L
=====
c = 0.087 06/0.120
c = 0.725 mol·L⁻¹
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
