Answer:
Sodium (Na): (.5 point)
22.99+35.453=58.433. 22.99/58.433= 39%
Chlorine (Cl): (.5 point)
22.99+35.453=58.433. 35.453/58.433= 61%
Explanation:
A asystem at equilibrium stops
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:
2Al + 3Ni(SO4) -------> 1Al2(SO4)3 + 2Ni
Explanation:
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