The grams of ethane present in a sample containing 0.4271 mole is 12.84 g
<h3>Description of mole </h3>
The mole of a substance is related to it's mass and molar mass according to the following equation
Mole = mass / molar mass
With the above formula, we can obtain the mass of ethane. Details below
<h3>How to determine the mass of ethane</h3>
The following data were obtained from the question:
- Mole of ethane = 0.4271 mole
- Molar mass of ethane = 30.067 g/mol
- Mass of ethane =?
The mass of ethane present in the sample can be obtained as follow:
Mole = mass / molar mass
Cross multiply
Mass = mole × molar mass
Mass of ethane = 0.4271 × 30.067
Mass of ethane = 12.84 g
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Answer:
1s22s22p63s1 is the electronic configuration of sodium.
Explanation:
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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.
3.44x10^2
you move the decimal over to get a single digit number with change. The number of times you move the decimal is the number for the 10 power
