Answer:
Sodium and chloride ions separate when salts are dissolved in water. The dissolved sodium and chloride ions, in high concentrations, can displace other mineral nutrients in the soil. Plants then absorb the chlorine and sodium instead of needed plant nutrients such as potassium and phosphorus, leading to deficiencies.
Explanation:
the explanation is given inyour answer.
Answer:
D) cutting the concentrations of both NOBr and NO in half
Explanation:
The equilibrium reaction given in the question is as follows -
2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )
The equilibrium constant for the above reaction can be written as -
K = [ NO ]² [ Br₂ ] / [ NOBr ] ²
Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,
Hence ,
the new concentrations are as follows -
[ NoBr ] ' = 1/2 [ NoBr ]
[ NO ] ' = 1/2 [ NO ]
Hence the new equilibrium constant equation can be written as -
K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²
Substituting the new concentration terms ,
K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²
K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²
The value of 1 / 4 in the numerator and the denominator is cancelled -
K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²
Hence ,
K' = K
Water, lower, and I'm not too sure for the third one... sorry.
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
I believe it’s C.) Mass. Hope I’m right.
