Answer:
The mass of the neon gas m = 1.214 kg
Explanation:
Pressure = 3 atm = 304 k pa
Volume = 0.57 L = 0.00057 
Temperature = 75 °c = 348 K
Universal gas constant = 0.0821 
We have to change the unit of this constant. it may be written as
Universal gas constant = 8.314 
Gas constant for neon =
= 0.41 
From ideal gas equation,
P V = m R T ------- (1)
We have all the variables except m. so we have to solve this equation for mass (m).
⇒ 304 ×
× 0.00057 = m × 0.41 × 348
⇒ 173.28 = 142.68 × m
⇒ m = 1.214 kg
This is the mass of the neon gas.
Glucose + Oxygen > Carbon Dioxide
C(6)H(12)0(6)+ 0(2) >CO(2)
The answer is A, to my brain power ig B)
Answer:
A. Whenever the population has increased, steel consumption has increased as well.
Explanation:
Based on the graph of US population and steel consumption, what could have led to the increase in steel consumption seen on the graph is that whenever the population has increased, steel consumption has increased as well.
A critical look at the graph, you will discover that the population and the steel consumption are moving upwards (i.e they are increasing). It's seen that as the population increases, steel consumption increases. This is true because as the population increases, people are building houses, more transportation systems that require steel are being manufactured, more household utensils that are steel products are being fabricated etc; therefore the consumption of steel increases.
When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M