Answer:
1. Hidracidas a. MX
2 Acidas c. MHXO
3. Oxacidas b. MXO
4. Basicas d. M(OH)X
Explanation:
¡Hola!
En este caso, de acuerdo con el concepto de sal, la cual está generalmente dada por la presencia de al menos un metal y un no metal, es posible encontrar cuatro tipos de estas; hidrácidas, oxácidas, básicas y ácidas, en las que las primeras dos son neutras pero la segunda tiene presencia de oxígeno, la tercera tiene iones hidróxido adicionales y la cuarta iones hidrógeno de más.
Debido a la anterior, es posible relacionar cada pareja de la siguiente manera:
1. Hidracidas a. MX
2 Acidas c. MHXO
3. Oxacidas b. MXO
4. Basicas d. M(OH)XO
En las que M se refiere a un metal, X a un no metal, H a hidrógeno y O a oxígeno.
¡Saludos!
Answer:
2.52L
Explanation:
Given parameters:
T₁ = 400K
V₁ = 4L
T₂ = 252K
unknown
V₂ = ?
Solution:
To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.
Mathematically,
Substitute and solve for V₂
V₂ = 2.52L
Answer:
The choice of the answer is fourth option that is -61 degrees.
Therefore the temperature drop is -61°Centigrade.
Explanation:
Given:
The temperature in a town started out at 55 degrees
Start temperature = 55°Centigrade. (Initial temperature)
End of the Day = -6°Centigrade. (Final temperature)
To Find:
How far did the temperature drop?
Solution:
We will have,
Substituting the above values in it we get
Therefore the temperature drop is -61°Centigrade.
Answer:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
E decreseas 3/2 as fast as G increases = 0.30 M/s
Explanation:
Rate of reaction = -d[D] / 2dt = -d[E]/ 3dt = -d[F]/dt = d[G]/2dt = d[H]/dt
When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:
Given data = d[D]/dt = 0.10 M/s
-d[D] / 2dt = d[H]/dt
d[H]/dt = 0.05 M/s
The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1
When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:
d[G] / 2dt = -d[H]/3dt
E decreseas 3/2 as fast as G increases = 0.30 M/s
