Explanation:
Because molarity is mol/L, we'll have to convert 17g to mol.
After obtaining the mol, we'll divide that by the volume to obtain Molarity.
Answer:
The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Explanation:
Water potential = Pressure potential + solute potential


We have :
C = 0.15 M, T = 273.15 K
i = 1
The water potential of a solution of 0.15 m sucrose= 
(At standard temperature)


The water potential of a solution of 0.15 M sucrose solution is -3.406 bar.
Answer:
- <u><em>1.12 liters</em></u>
Explanation:
<u>Calculating number of moles</u>
- Molar mass of O₂ = 32 g
- n = Given weight / Molar mass
- n = 1.6/32
- n = 0.05 moles
<u>At STP</u>
- One mole of O₂ occupies 22.4 L
- Therefore, 0.05 moles will occupy :
- 22.4 L x 0.05 = <u><em>1.12 L</em></u>
Answer:
0.007 M
Explanation:
pH is defined as the negative logarithm of the concentration of hydrogen ions.
Thus,
pH = - log [H⁺]
The expression of the pH of the calculation of weak acid is:-
Where, C is the concentration = ?
Given, pH = 3.45
So, for
,

C = 0.007 M
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol