Answer:
option D is correct
D. This solution is a good buffer.
Explanation:
TRIS (HOCH
)
CNH
if TRIS is react with HCL it will form salt
(HOCH)CNH + HCL ⇆ (HOCH)NHCL
Let the reference volume is 100
Mole of TRIS is = 100 × 0.2 = 20
Mole of HCL is = 100 × 0.1 = 10
In the reaction all of the HCL will Consumed,10 moles of the salt will form
and 10 mole of TRIS will left
hence , Final product will be salt +TRIS(9 base)
H = Pk
+ log (base/ acid)
8.3 + log(10/10)
8.3
Answer:
k = -0.006.
T₀ = 15 °C
Explanation:
Hola.
En este caso, considerando la gráfica mostrada en el archivo adjunto, podemos evidenciar que los datos dados se comportan de manera lineal, por lo que basado en la ecuación, T=k*h+To, podemos calcular la pendiente que basicamente es igual a k, tomando dos puntos en la gráfica:
Además, el valor de la temperatura inicial se puede extraer de la tabla, dado que esta es cuando la altura es 0 m, es decir 15 °C.
¡Saludos!
Answer:
Option b. Decomposition
Followed by a reduction process using charcoal
Explanation:
Lead can be obtained from lead nitrate by thermal decomposition of lead nitrate as shown below:
2Pb(NO3)2 —> 2PbO + 4NO2 + O2
The PbO obtained is reduced by charcoal(C) to obtain the metallic Pb as shown below:
2PbO + C —> Pb + CO2
C.) remove yourself from the radiation every 15 mins to prevent accumulation
You start by using proportions to find the number of liters of solution:
180 g of glucose / 1 liter of solution = 18 g of glucose / x liter of solution
=> x = 18 g of glucose * 1 liter of solution / 180 g of glucose = 0.1 liter of solution.
If you assume that the 18 grams of glucose does not apport volume to the solution but that the volume of the solution is the same volumen of water added (which is the best assumption you can do given that you do not know the how much the 18 g of glucose affect the volume of the solution) then you should add 0.1 liter of water.
Answer: 0.1 liter of water.
