Answer:
We need 7.5 mL of the 1M stock of NaCl
Explanation:
Data given:
Stock = 1M this means 1 mol/ L
A 0.15 M solution of 50 mL has 0.0075 moles NaCl per 50 mL
Step 2: Calculate the volume of stock we need
The moles of solute will be constant
and n = M*V
M1*V1 = M2*V2
⇒ with M1 = the initial molair concentration = 1M
⇒ with V1 = the volume we need of the stock
⇒ with V2 = the volume we want to make of the new solution = 50 mL = 0.05 L
⇒ with M2 = the concentration of the new solution = 0.15 M
1*V1 = 0.15*(50)
V1 = 7.5 mL
Since 0.0075 L of 1M solution contains 0.0075 moles
50 mL solution will contain also 0.0075 moles but will have a molair concentration of 0.0075 moles / 0.05 L =0.15 M
We need 7.5 mL of the 1M stock of NaCl
PH of solution will be greater than seven (pH>7), that means that solution is basic (<span>pH above </span>7<span> is a base, the higher the number, the stronger is the base).
</span>pH (potential of hydrogenis) is a measure of the hydrogen ion (H⁺) concentration of a solution. <span>Solutions with a pH less than 7 are acidic.</span>
Explanation:
is used for making ropes, used for climbing rocks and for making parachutes. Their usage shows that nylon fibres have high tensile strength
Answer:
Water would not be able to transport nutrients -‐-‐ in plants, or in our bodies -‐-‐ nor to dissolve and transport waste products out of our bodies. ... Cohesiveness, adhesiveness, and surface tension: would decrease because without the +/-‐ polarity, water would not form hydrogen bonds between H20 molecules.
Answer:
The new partial pressures after equilibrium is reestablished:
Explanation:
At equilibrium before adding chlorine gas:
Partial pressure of the 
Partial pressure of the 
Partial pressure of the 
The expression of an equilibrium constant is given by :
At equilibrium after adding chlorine gas:
Partial pressure of the 
Partial pressure of the 
Partial pressure of the 
Total pressure of the system = P = 263.0 Torr
At initail
(13.2) Torr (32.8) Torr (13.2) Torr
At equilbriumm
(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr
Solving for x;
x = 6.402 Torr
The new partial pressures after equilibrium is reestablished:
