Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
We can calculate the final temperature from this formula :
when Tf = (V1* T1) +(V2* T2) / (V1+ V2)
when V1 is the first volume of water = 5 L
and V2 is the second volume of water = 60 L
and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K
and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K
and Tf is the final temperature of water in Kelvin
so, by substitution:
Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)
= 1765 + 18180 / 65 L
= 306 K
= 306 -273 = 33° C
I’m not 100% positive but quartz would make sense
First, we convert the pressure in kPa to atm. That is,
(215 kPa) x (1 atm/ 101.325 kPa) = 2.122 atm
Then, determine the amount of carbon (C) in mol by assuming that it is an ideal gas.
n = (2.122 atm)(75 L) / (0.0821 L.atm/mol.K)(-25 + 273.15)
n = 7.81 mols
The reaction is expressed as,
4KO2 + 2CO2 --> 2K2CO3 + 3O2
From the equation,
(7.81 mols CO2) x (4 mols KO2/2 mols CO2) = 15.62 mols KO2
Answer : 70.906 grams of chlorine reacted with hydrogen
Explanation :
Step 1 : Write balanced chemical equation.
The balanced chemical equation for the reaction between hydrogen and chlorine gas is given below.
Step 2 : Find moles of H₂ gas.
The moles of H₂ can be found as
We have 2.0200 g of H₂ and molar mass of H₂ is 2.02 g/mol.
Let us plug in these values to find moles of H₂.
We have 1 mol of H₂.
Step 3 : Find moles of Cl₂ using mole ratio.
The mole ratio of H₂ and Cl₂ is 1 : 1.
The moles of Cl₂ can be calculated as
Step 4 : Find grams of Cl₂.
Molar mass of Cl₂ gas is 70.096 g/mol
Mass of Cl₂ = 
We have 70.906 grams of Cl₂
