Answer:
We need 78.9 mL of the 19.0 M NaOH solution
Explanation:
Step 1: Data given
Molarity of the original NaOH solution = 19.0 M
Molarity of the NaOH solution we want to prepare = 3.0 M
Volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
Step 2: Calculate volume of the 19.0 M NaOH solution needed
C1*V1 = C2*V2
⇒with C1 = the concentration of the original NaOH solution = 19.0 M
⇒with V1 = the volume of the original NaOH solution = TO BE DETERMINED
⇒with C2 = the concentration of the NaOH solution we want to prepare = 3.0 M
⇒with V2 = the volume of the NaOH solution we want to prepare = 500 mL = 0.500 L
19.0 M * V2 = 3.0 M * 0.500 L
V2 = (3.0 M * 0.500L) / 19.0 M
V2 = 0.0789 L
We need 0.0789 L
This is 0.0789 * 10^3 mL = 78.9 mL
We need 78.9 mL of the 19.0 M NaOH solution
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:E
Explanation:
The electron is expected to be in any of the 4p orbitals. For the 4p orbitals
n=4
l=1
ml=-1,0,1
ms=-1/2 or +1/2
All the ml and ms configurations are equiprobable hence the answer.
One gram of sodium chloride takes longer to dissolve in a solution of 2M glucose than in pure water. This is because diffusion happens faster in pure water than in a solution containing a solute i.e. glucose and thus it will take longer to dissolve as a result of large number of collisions.
If you were to put 1 gram of sodium chloride (NaCl) in a beaker of pure water, it would eventually dissolve. However, if you dissolve same amount of sodium chloride (NaCl) in a beaker of water containing 2 moles of glucose, sodium chloride would take longer to completely dissolve.
The reason for this is that, in a solution having more molecules, the sodium chloride would experience more collisions. Due to this, the rate of diffusion will slow down, and thus, sodium chloride will take longer to completely dissolve.
To put it simply, solutions having large number of molecules present exhibit slower diffusion rate than solutions with fewer molecules present. So, if you want the sodium chloride to dissolve quickly, the best way is to put it in a solution of pure water.
If you need to learn more about sodium chloride click here
brainly.com/question/24878544
#SPJ4
