Explanation:
I don't know sorry ok I have some work so I m sorry
Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
Commonly, sacrificial electrodes are employed to stop another metal from corroding or oxidising. A metal that is more reactive than the metal being shielded must serve as the sacrificial electrode. Magnesium, aluminium, and zinc are the three metals most frequently used in sacrificial anodes.
Manganese-Magnesium (Mn-Mg) electrode is more suited for on-shore pipelines where the electrolyte (soil or water) resistivity is higher since it has the highest negative electropotential of the three. In order to replenish any electrons that could have been lost during the oxidation of the shielded metal, the highly active metal offers its electrons.
Therefore, Mn metal can be used as a sacrificial electrode to prevent the rusting of an iron pipe. So, the correct option is (c) Mn.
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Answer:
1.8 moles of NaCl must be produced.
Explanation:
Based on the reaction:
HCl + NaOH → NaCl + H2O
<em>1 mol of HCl reacts with 1 mol of NaOH to produce 1mol of NaCl</em>
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To solve this question we must find, as first, the <em>limiting reactant:</em>
<em />
1.8 moles of HCl will need 1.8 moles of NaOH for a complete reaction (Ratio of reaction 1:1). As there are 3.3 moles of NaOH,
<em>HCl is limiting reactant</em>
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When the 1.8 moles of HCl react completely,
1.8 moles of NaCl must be produced because 1 mole of HCl produce 1 mole of NaCl
1.34 L of HF
Explanation:
We have the following chemical reaction:
Sn (s) + 2 HF (g) → SnF₂ (s) + H₂ (g)
First we calculate the number of moles of SnF₂:
number of moles = mass / molecular weight
number of moles of SnF₂ = 5 / 157 = 0.03 moles
From the chemical reaction we see that 1 mole of SnF₂ are produced from 2 moles of SnF₂. This will mean that 0.03 moles of SnF₂ are produced from 0.06 moles of HF.
Now at standard temperature and pressure (STP) we can use the following formula to calculate the volume of HF:
number of moles = volume / 22.4 (L/mole)
volume of HF = number of moles × 22.4
volume of HF = 0.06 × 22.4 = 1.34 L
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