Answer:
1. molarity is = 0.1174
2. molarity is = 9.3
3. molarity is =0.009392
Explanation:
Q#1.
mass of sucrose is 14.07g
molar mass of sucrose is 342 3g/mol
so, 14.07x1/342.3 = 0.0411
now the volume is of water is 0.35
so, 0.0411/0.35 = 0.1174
Q#2.
concentration is 4.5
so, 4.5x1/1000= 0.0045
then, 0.0411/0.0045 = 9.13
Q#3.
M1M1=M2M2
molarity of the same sucrose is 0.1174
so, diluted in 40mL
and the volumetric flask is 500mL
so, 0.1174x40/500= 0.009392
Answer:
A
Explanation:
because It's the process by which atmospheric nitrogen is converted either by a natural or an industrial means to form of nitrogen such as ammonia.
Answer:
The value of the missing equilibrium constant ( of the first equation) is 1.72
Explanation:
First equation: 2A + B ↔ A2B Kc = TO BE DETERMINED
⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]
Second equation: A2B + B ↔ A2B2 Kc= 16.4
⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]
Third equation: 2A + 2B ↔ A2B2 Kc = 28.2
⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²
If we add the first to the second equation
2A + B + B ↔ A2B2 the equilibrium constant Kc will be X(16.4)
But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2) with Kc = 28.2
So this means: 28.2 = X(16.4)
or X = 28.2/16.4
X = 1.72
with X = Kc of the first equation
The value of the missing equilibrium constant ( of the first equation) is 1.72
Answer:
control.
Explanation:
during an experiment you are required to maintain a separate group of subjects to collect data on so you will be able to make comparisons from your observations. assuming the watered plants grew, what does that mean? they grew at a quicker rate? slower rate? the same rate? compared to what? you need this control group in order to prove your observations either one way or the other such as "compared to unwatered plants, the watered plants grew at *blank* rate."
Answer:
Water vapor has mostly disappeared at colder elevations and its effect on weather turbulence.
Explanation:
