Answer:
The empirical formula of compound is C₂H₆O.
Explanation:
Given data:
Mass of carbon = 12 g
Mass of hydrogen = 3 g
Mass of oxygen = 8 g
Empirical formula of compound = ?
Solution:
First of all we will calculate the gram atom of each elements.
no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms
no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms
no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms
Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.
C:H:O = 1/0.5 : 3/0.5 : 0.5/0.5
C:H:O = 2 : 6 : 1
The empirical formula of compound will be C₂H₆O
Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:

From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:

= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:

= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:

= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:

= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
2: <span>Volume V = a*b*c = 6.0*3.0*3.0 = 54.0 cm^3 density ρ = mass/volume = 146/54 = 2.70 g/cm^3
3: Volume = (27.8 -21.2) cm^3
mass = 22.4 g
density = 22.4/(27.8-21.2) g/cm^3
</span>
Answer:
The correct answer is 0.300 * 10^23 ions.
Explanation:
Based on the given question, there is a need to find the number of chloride ions in the mentioned 6.8 grams of zinc chloride compound.
The moles of zinc chloride (ZnCl2) is,
= mass of zinc + 2 mass of chlorine
= 65.38 + 2 (35.45)
=65.38 + 70.90
= 136.28 grams (The molecular mass of zinc is 65.38 and the molecular mass of chlorine is 35.45)
Thus, 136.28 g of ZnCl2 contains 70.90 grams of chlorine
Therefore, 6.8 grams of ZnCl2 will comprise = (70.90/136.28) * 6.8
= 3.537 g of chlorine
70.90 g of Cl comprise 6.022*10^23 chlorine, thus, 3.537 g of Cl will comprise (6.022*10^23/70.90) * 3.537
= 0.300 * 10^23 ions of chlorine.
Answer : The pH of 0.289 M solution of lithium acetate at
is 9.1
Explanation :
First we have to calculate the value of
.
As we know that,

where,
= dissociation constant of an acid = 
= dissociation constant of a base = ?
= dissociation constant of water = 
Now put all the given values in the above expression, we get the dissociation constant of a base.


Now we have to calculate the concentration of hydroxide ion.
Formula used :
![[OH^-]=(K_b\times C)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%28K_b%5Ctimes%20C%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
where,
C is the concentration of solution.
Now put all the given values in this formula, we get:
![[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%285.5%5Ctimes%2010%5E%7B-10%7D%5Ctimes%200.289%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
![[OH^-]=1.3\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.3%5Ctimes%2010%5E%7B-5%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 of 0.289 M solution of lithium acetate at
is 9.1