The answer to this question would be: lower molar concentration
Osmotic pressure is influenced by the number of ions and the concentration of the molecule in the solution. In NaCl, the molecule will split into 1 Na+ ion and 1 Cl- ion which results in 2 ions per compound. In MgCl2, the compound will split into 1 Mg2+ ion and 2 Cl- ion which results in 3 ions. Therefore, the osmotic pressure of MgCl2 will be 3/2 times of NaCl.
MgCl2 will need less concentration to achieve same osmotic pressure as NaCl. If the MgCl2 solution is isotonic with NaCl, the concentration of MgCl2 would be lower than NaCl
Answer : The moles of 
are, 2.125 mole.
Explanation : Given,
Molarity of = 8.500 M
Volume of solution = 250 mL = 0.250 L (1 L = 1000 mL)
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
Now put all the given values in this formula, we get:
Therefore, the moles of are, 2.125 mole.
A human body, just like a dog's, will sweat. Dogs will pants and sweat through the pads of their feet to cool down, and human will sweat through their foreheads, armpits, etc.
Dogs will tend to, in hot environments, lay on the floor or where the surface is cooler. Since they cannot simply strip their clothing to keep cool they tend to find cool surfaces, fans, sources of air, etc. to keep cool from the heat.
Answer is: <span>the molarity of the diluted solution 0,043 M.
</span>V(NaOH) = 75 mL ÷ 1000 mL/L = 0,075 L.
c(NaOH) = 0,315 M = 0,315 mol/L.
n(NaOH) = c(NaOH) · V(NaOH).
n(NaOH) = 0,075 L · 0,315 mol/L.
n(NaOH) = 0,023625 mol.
V(solution) = 0,475 L + 0,75 L.
c(solution) = 0,023625 mol ÷ 0,550 L.
c(solution) = 0,043 mol/L.
Answer:
3.676 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
- Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
