C. alpha = nucleus of a helium atom
beta = electron
gamma = photon
Answer:
H2SO4 + 8HI → H2S + 4I2 + 4H2O
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
<u>Answer:</u> The number of
ions dissociated are
<u>Explanation:</u>
We are given:
pH = 2.07
Calculating the value of pOH by using equation, we get:

To calculate hydroxide ion concentration, we use the equation to calculate pOH of the solution, which is:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
pOH = 11.93
Putting values in above equation, we get:
![11.93=-\log[OH^-]](https://tex.z-dn.net/?f=11.93%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-11.93}=1.17\times 10^{-12}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-11.93%7D%3D1.17%5Ctimes%2010%5E%7B-12%7DM)
To calculate the number of moles for given molarity, we use the equation:

Molarity of solution = 
Volume of solution = 1243 mL = 1.243 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:

According to mole concept:
1 mole of a compound contains
number of particles
So,
number of
will contain =
number of ions
Hence, the number of
ions dissociated are
Sorry but i can not decide i think it is B or C