Answer:
V = 22.34 L
Explanation:
Given data:
Volume of O₂ needed = ?
Temperature and pressure = standard
Number of molecules of water produced = 6.0× 10²³
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of water:
1 mole contain 6.022× 10²³ molecules
6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules
0.99 mole
Now we will compare the moles of oxygen and water.
H₂O : O₂
2 : 1
0.996 : 0.996
Volume of oxygen needed:
PV = nRT
V = nRT/P
V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm
V = 22.34 L
The balanced chemical reaction would be:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:
Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl
0.646 - 0.48 = 0.166 mol HCl left
The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.
The balanced chemical equation for the reaction can be represented as,
Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL
Heat of the reaction, q = 
Δ
m is mass of the solution = 151.8 mL * 
C is the specific heat of solution = 4.18 
ΔT is the temperature change = 
q = 
Moles of NaOH = 
NaOH
Moles of 
= 
Enthalpy of the reaction = 
Answer:
Exergonic ,Endergonic,low concentration area,high
Explanation:
In exergonic reaction,certain molecules are broken down;in the process they release energy which is captured when high energy molecules(such as ATP and NADH) are formed.
The breakdown of these molecules can be coupled to thermodynamically unfavorable processes such as Endergonic reactions or pumping og hydrogen ion from low concentration areas to high concentration areas.
Explanation:
Reduction is a chemical reaction in which electrons are gained by one of the atoms taking part in the reaction and lowering of an oxidation state of that atom.
Reduction takes place at the cathode.
In aqueous, vanadium(V) is present in +5 oxidation state which on reduction changes to vanadium(I) with +1 oxidation state.
The half reaction is :
