I think the effect of increasing temperature would be; the equilbrium will shift back wards. Increase in temperature favors backward reaction since the forward reaction is exothermic and the backward reaction is endothermic. Therefore, the equilibrium will shift back wards, and there will be more reactants (H2 and Cl2) compared to the products
Answer:
Air and Water Temperature Increases
An increase in the air temperature will cause water temperatures to increase as well. ... Lower levels of dissolved oxygen due to the inverse relationship that exists between dissolved oxygen and temperature. As the temperature of the water increases, dissolved oxygen levels decrease.
The values of the coefficients would be 4, 5, 4, and 6 respectively.
<h3>Balancing chemical equations</h3>
The equation of the reaction can be represented by the following chemical equation:
ammonia (g) + oxygen (g) ---> nitrogen monoxide (g) + water (g)
+ 
---> 
+ 
Thus, the coefficient of ammonia will be 4, that of oxygen will be 5, that of nitrogen monoxide will be 4, and that of water will be 6.
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Answer:
state of matter
Explanation:
so take water for example, water has a melting point and a boiling point right? So if it's below 0 degrees, then it's in its solid phase. If the temperature is above 0 degrees, then the water starts to melt into its liquid phase. Then when the temperature is above 100 degrees, water starts to boil and become its gas phase. This is the same for all substances. The only difference is different substances have different melting and boiling points so the numbers will be different depending on your substance. hope this helped!
The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
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