To determine the mass of sucrose from a given volume of solution, we need to convert the volume into mass by using the density of the solution. We calculate as follows:
mass solution = 3.50 ( 1118 ) = 3913 g
mass of sucrose = 3913 g solution ( .485 g sucrose / g solution ) = 1897.805 g sucrose is present in the solution.
Answer:
c solvents dissolve chemicals with the same polarity ( ex. both are polar)
Explanation:
Like dissolves like is one of the central rule that guides the solubility of one substance in another.
- It fully suggests substance having the same nature as in polarity-wise will dissolve one another.
- For example, water is a polar liquid, it will dissolve table salt because it i also polar.
- Water cannot dissolve oil because oil is non-polar.
Answer:
C₃H₉N
Explanation:
The empirical formula of a compound is the fundamental and basic possible formula that shows the mole ratio of the atoms of each element in a molecule of the compound.
mole ratio of carbon = 60.94/12 = 5.078
mole ratio of hydrogen = 15.36/1 = 15.36
mole ratio of nitrogen = 23.70/14 = 1.693
Now; we will divide by the smallest value
So; carbon = 5.078/1.693 = 2.99 ≅ 3.0
hydrogen = 15.36/1.693 = 9.07 ≅ 9.0
nitrogen = 1.693/1.693 = 1 ≅ 1
Thus, the empirical formula is = C₃H₉N
Answer: C2H4
Explanation:
The percentage composition of ethanol ( C2H5OH ) consist of 52.2% Carbon, Hydrogen of 13.0 and 34.8% of Oxygen.
The percentage composition of ethane gas (C2H6) consist of 80.0% carbon and 20.0% hydrogen.
The composition of Ethylene Glycols i.e C2H4(OH)2 is Carbon of 39.7%, 9.7% hydrogen and 51.6% oxygen.
The percent composition of c2h4 is 86% carbon, and 14% hydrogen.
From the information given, the substance with the highest percentage of carbon is C2H4
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is 
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 
Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
