Answer:
Twice as much.
Explanation:
That's because the freezing point depression depends on the total number of solute particles.
C₆H₁₂O₆(s) ⟶ C₆H₁₂O₆(aq)
0.01 mol of C₆H₁₂O₆ gives 0.01 mol of solute particles.
NaCl(s) ⟶ Na⁺(aq) + Cl⁻(aq)
1 mol of NaCl gives 0.01 mol of Na⁺(aq) and 0.01 mol of Cl⁻(aq).
That's 0.02 mol of particles, so the freezing point depression of 0.01 mol·L⁻¹ NaCl will be twice that of 0.01 mol·L⁻¹ C₆H₁₂O₆.
Answer : The pH will be, 3.2
Explanation :
As we known that the value of solubility constant of ferric hydroxide at
is, 
Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L
The given solubility of iron convert from mg/L to mol/L.

The chemical reaction will be:

The expression of solubility constant will be:
![K_{sp}=[Fe^{3+}]\times [3OH^-]^3](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B3%2B%7D%5D%5Ctimes%20%5B3OH%5E-%5D%5E3)
Now put all the given values in this expression, we get the concentration of hydroxide ion.
![2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3](https://tex.z-dn.net/?f=2.79%5Ctimes%2010%5E%7B-39%7D%3D%282.7%5Ctimes%2010%5E%7B-7%7D%29%5Ctimes%20%5B3OH%5E-%5D%5E3)
![[OH^-]=1.5\times 10^{-11}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.5%5Ctimes%2010%5E%7B-11%7DM)
Now we have to calculate the pOH.
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)


Now we have to calculate the pH.

Therefore, the pH will be, 3.2
The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
answer: 3.40625 moles
Explanation: see the attached pics
Explanation: We are given three balloon carrying different charges: One having positive charge, one having negative charge and one having no charge.
When we bring positive rod near these three balloons, all of them behave differently.
According to the Coulomb's Law:
Like charges repel each other and unlike charges attract each other.
- When we bring positive rod towards the balloon having positive charge, the balloon will move away from the rod, because like charges repel each other.
- When we bring positive rod near negative balloon, the balloon will be attracted to the rod because unlike charges attract each other.
- But, when we bring positive rod towards the balloon having no charge, the charges will be induced in the balloon. Although the balloon is neutral in nature but it will still carry some +ve or -ve charges. So, when the positive rod is bought near neutral ball, the negative side of the balloon will get attracted towards the rod and positive side will be repelled. From this, we say that the rod has induced some charges in the balloon.