The amount, in grams, of ethanol needed to prepare 268 grams of a 9.76 M solution of ethanol in water will be 120.50 grams
<h3>Solution preparation</h3>
In order to prepare 268 grams of solution, 268 mL of water would be needed because 1 mL of water weighs 1 gram.
Since we now know the volume of solution that we want to prepare, the amount of solute (ethanol) that would be required in order to make a solution of 9.76 M will be:
Mole required = 9.76 x 268/1000 = 2.62 moles
Mass of 2.62 moles ethanol = 2.62 x 46.07 = 120.50 grams
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It is a homogenous mixture because when iron and rust mix it’s homogenous.
We know, by law of constant proportion :
Chemical compounds always contains its component in constant fixed ratio or in fixed percentage without depend on its source, method of preparation and mass of compound.
Therefore, percentage of oxygen is same for any mass.
Hence, this is the required solution.
Across the period
Explanation:
Among the element of the main group, the first ionization energy increases across the period.
- The energy required to remove an electron from an atom is called ionization energy.
- The first ionization energy is the energy needed to remove the most loosely bound electron of an atom in the gas phase.
- The ionization energy measures the readiness of an atom to lose an electron.
The magnitude of the ionization energy depends on nuclear charge, atomic radius, sublevel accomodating the electron and the special stability of filled and half-filled sublevels.
Generally, ionization energy increases progressively from left to right due to the decreasing atomic radii.
Down a group, it decreases because atomic radii is increasing
Learn more:
Ionization energy brainly.com/question/1971327
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