Air Pressure drops more rapidly with altitude in a column of cold air than in warm air.The answers to this question are cold air and warm air, respectively.
<span>Cold air is known to be dense while warm air is known otherwise to be less thens which makes it move upwards. Cold air experiences more pressure as it moves upwards.</span>
Balanced chemical reaction:
2Na₃PO₄(aq) + 3CaCl₂(aq) → 6NaCl(aq) + Ca₃(PO₄)₂(s).
Ionic reaction:
6Na⁺(aq) + 2PO₄³⁻(aq) + 3Ca²⁺(aq) + 6Cl⁻(aq) → 6Na⁺(aq) + 6Cl⁻(aq) + Ca₃(PO₄)₂(s).
Net ionic reaction: 2PO₄³⁻(aq) + 3Ca²⁺(aq) → Ca₃(PO₄)₂(s).
<span>(aq) means that
substances are dissociated on cations and anions in water.
</span>(s) means solid.
Pasteur said that heating the wine at about 120°F would make sure that the wine would not turn sour, so the option is A.
Here's the information link: http://www.faqs.org/health/bios/83/Louis-Pasteur.html
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.
Learn more about electrode here:
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Answer: Option D) 0.902M
Explanation:
Sodium chloride has a chemical formula of NaCl
Given that,
Amount of moles of NaCl (n) = ?
Mass of NaCl in grams = 145g
For molar mass of NaCl, use the molar masses:
Sodium, Na = 23g;
Chlorine,Cl = 35.5g
NaOH = (23g + 35.5g)
= 58.5g/mol
Since, amount of moles = mass in grams / molar mass
n = 145g / 58.5g/mol
n = 2.48 mole
Amount of moles of NaCl (n) = 2.48 mole
Volume of NaCl solution (v) = 2.75L
Concentration of NaCl solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 2.48 mole / 2.75L
c = 0.9018M
Approximately, c = 0.902M
[c is the concentration in moles per litres which is also known as molarity]
Thus, the molarity of the solution is 0.902M