Answer:
Oxidation half-reaction : Na(s) → Na⁺ + 1 e-
Reduction half-reaction: Cl₂ + 2 e- → 2 Cl⁻
Explanation:
Oxidation half-reaction: solid sodium (Na(s)) has an oxidation number of 0. It loses 1 electron and forms the cation Na⁺. So, the half-reaction is the following:
Na(s) → Na⁺ + 1 e-
Reduction half-reaction: chlorine gas (Cl₂) has an oxidation number of 0. Each atom of Cl gains 1 electron to form two Cl⁻ ions, according to the following half-reaction:
Cl₂ + 2 e- → 2 Cl⁻
The total oxidation-reduction reaction is obtained by adding the oxidation half-reaction multiplied by 2 (to balance the electrons) and the reduction half-reaction, as follows:
2 x (Na(s) → Na⁺ + 1 e-)
Cl₂ + 2 e- → 2 Cl⁻
--------------------------------
2Na(s) + Cl₂ → 2NaCl
Answer:
the answer is According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge. The electrons are like plums in a pudding.
Explanation:
Main Answer:
Given data:
Initial Pressure P1 = 570 mm hg
Initial Volume V1 = 2270 ml
Final Pressure P2 = ? mm hg
Final Volume V1 = 1250 ml
According to the ideal gas equation,
PV = constant.
P1V1 = P2V2
P2 = P1V1/V2
P2 = (570 x 2270) / 1250
P2 = 1035.12 mm hg
The final pressure at volume of 1250 ml is 1035.12 mm hg.
Explanation:
What is ideal gas equation ?
The ideal gas equation is as follows:
PV = nRT
where P = Pressure
V = Volume
n = number of moles of gas
R = Universal gas constant
T = Temperature
This ideal gas equation provides the macroscopic particles behavior of the gas. At this condition, the particles of the gas, won't be attract or repel each other. It is consider as the stable condition.
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Answer:
Buffers are resistant to high pH changes.
Explanation:
This perfectly explains the reason why we use buffers. Buffers are substances which consist of a weak acid and its conjugate base. Buffers are resistant to significant pH changes upon addition of strong acids or bases. To illustrate this, let's say we have a buffer consisting of 0.1 mol of HF, a weak acid, and 0.1 mol of NaF (fluoride is a conjugate base of HF).
- Let's say we add some strong acid, in a general form, this acid would be represented as . In this case, conjugate base will react and neutralize it to produce some amount of HF: .
- Similarly, if we add some strong base , the acidic component will react with it to produce some amount of conjugate base: . The ratio of HF to NaF in this case is held around the same value for addition of small amounts of strong acids/bases, so pH is kept almost constant, while in neutral water, pH would drastically increase or decrease.