P₄0₆
+ 20₂
⇒ P₄0₁₀
Explanation:
The overall equation for the reaction that produces P₄0₁₀ is :
P₄0₆
+ 20₂
⇒ P₄0₁₀
Now let us derive this equation:
Given equations:
P₄
+ 30₂
⇒ P₄0₆
equation 1;
P₄
+ 50₂
⇒ P₄0₁₀
equation 2;
To get the overall combined equation, the equation 1 must be reversed and added to equation 2:
P₄0₆
⇒ P₄
+ 30₂
equation 3
+
equation 2:
P₄
+ 50₂
+ P₄0₆
⇒ P₄0₁₀
+ P₄
+ 30₂
cancelling specie that appears on both sides and removing excess oxygen gas on the reactant side gives;
P₄0₆
+ 20₂
⇒ P₄0₁₀
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Oxidation is the half reaction that can occur at the anode in a voltaic cell.
Explanation:
In electrodes which is metal strip in voltaic cell the reactions occurs. The two electrodes placed one in each half-cell. The reduction reaction occurs at cathode and oxidation occurs at anode.
A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. Half-reactions are often used as a method of balancing redox reactions.
In an oxidation half reaction, an atom loses electron(s). When an element is oxidized it loses a specific number of electrons.
The gas inside the can and the can’s volume are both constant.
The gas pressure increases with increasing temperature.
The can will burst if the pressure becomes great enough.
The gas law that applies is Gay-Lussac’s law.
Answer:
At 0.58 L of 0.540 M NaOH solution contain 12.5 g NaOH.
Explanation:
Given data:
At volume = ?
Mass of NaOH = 12.5 g
Molarity of solution = 0.540 M
Solution:
First of all we will calculate the number of moles of sodium hydroxide.
Number of moles = mass/molar mass
Number of moles = 12.5 g / 40 g/mol
Number of moles = 0.3125 mol
Volume of NaOH:
Molarity = number of moles / volume in L
Now we will put the values.
0.540 M = 0.3125 mol / volume in L
volume in L = 0.3125 mol / 0.540 mol/L
volume in L = 0.58 L