Answer:
As soon as a front passes a particular location, the weather could change from a steamy maritime tropical air mass to a dry.
Explanation:
hope this helps
When we can get the Kinetic energy from this formula KE= 1/2 M V^2
and we can get the potential energy from this formula PE = M g H
we can set that the kinetic energy at the bottom of the fall equals the potential energy at the top so,
KE = PE
1/2 MV^2 = M g H
1/2 V^2 = g H
when V is the velocity, g is an acceleration of gravitational force and H is the height of the fall.
∴ v^2 = 2 * 9.8 * 8 = 156.8
∴ v= √156.8 = 12.5 m/s
Answer : The correct answer for change in freezing point = 1.69 ° C
Freezing point depression :
It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .
SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .
It can be expressed as :
ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m
Where : ΔTf = change in freezing point (°C)
i = Von't Hoff factor
kf =molal freezing point depression constant of solvent.
m = molality of solute (m or 
)
Given : kf = 1.86 
m = 0.907 )
Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1
Plugging value in expression :
ΔTf = 1* 1.86 * 0.907 )
ΔTf = 1.69 ° C
Hence change in freezing point = 1.69 °C
Los alquimistas querían cambiar el plomo en oro. Esto es, para crear algo de nada
Answer:
Ka = ( About ) 5 x 10^ - 8
Explanation:
Let us first identify the dissociation equation for this weak acid,
HA ⇌ ( H+ ) + A¯
Knowing this, we can tell what the equilibrium expression is, respectively,
Ka = ( [ H+ ] [ A¯ ] ) / [ HA ]
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Now let us use the given pH 3.99 to calculate [ H+ ], knowing that pH = −log [H+],
3.99 = - log[H+],
[H+] = 10 ^ - 3.99,
[H+] = ( About ) 1 * 10^-4 M
Substitute known values into the equilibrium expression,
Ka = [( 1 x 10^ - 4 ) ( 1 x 10^ ¯4 )] / 0.199,
Ka = ( About ) 5 x 10^ - 8
