Answer:

Explanation:
k stand for equilibrium constants in terms of reaction
The higher the value of an equilibrium constant the faster the equilibrium reaction comes to completion.
Consider the example below:
⇄
where

For a faster reaction the numerator i.e. the right hand side of the equation have to be higher than the left hand side (the denominator). therefore the higher the numerator, the higher the value of the equilibrium constant and the faster the reaction get to completion thus option c is correct.
Carbon dioxide and oxygen are removed from the air.
Explanation:
When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.
First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :
CO₂ + H₂O → H₂CO₃
The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):
H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O
After this by passing the air over heated cooper the oxygen is removed.
2 Cu + O₂ → 2 CuO
Learn more about:
neutralization reaction
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Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
The number of protons would be equivalent to the number of electrons if the net charge on the atom is 0.
Answer:
Explained
Explanation:
The bond angle between the Si-O-Si is 145° .But this can also vary between 100° and 170°, It is because of which the changes in properties is happening. Bond angles can affect the hybridization, lone pair repulsion and electronegativity.
Thus , Space shuttle tiles, sandstone, and the glass in quartz have different properties and are used according to their property.