To determine the formula units of the given compound, we need to convert from grams to moles first then convert it to formula units by using the Avogadro's number. We do as follows:
23.5 g <span>Sb2S3 ( 1 mol / 339.7 g ) ( 6.022 x 10^23 formula units / mol ) = 4.17x10^22 formula units </span><span>Sb2S3</span>
<h3>Answer:</h3>
2.55 × 10²² Na Atoms
<h3>Solution:</h3>
Data Given:
M.Mass of Na = 23 g.mol⁻¹
Mass of Na = 973 mg = 0.973 g
# of Na Atoms = ??
Step 1: Calculate Moles of Na as:
Moles = Mass ÷ M.Mass
Moles = 0.973 g ÷ 23 g.mol⁻¹
Moles = 0.0423 mol
Step 2: Calculate No, of Na Atoms as:
As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,
Moles = No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹
Solving for No. of Na Atoms,
No. of Na Atoms = Moles × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 2.55 × 10²² Na Atoms
<h3>Conclusion: </h3>
2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.
Answer:
The answer to your question is: letter B.
Explanation:
Carbon atom double-bonded to an oxygen atom C= O,
a) Carbohydrate group this group does not exist
b)Carbonyl group This option is right, carbonyl group is C=O
c) Hydroxyl group This option is incorrect, hydroxyl group is OH, very different from C=O
d) Ketone group Ketone group does not exist, ketones have the same functional group as aldehydes (C=O)
The concentration of a solution of 192 grams dissolved to make 2150 mL of solution is 0.93M. Details about concentration can be found below.
<h3>How to calculate concentration?</h3>
The concentration of a solution can be calculated by dividing the number of moles of the substance by its volume.
no of moles of NH42CO3 = 192g ÷ 96g/mol = 2 mol
Concentration of solution = 2 moles ÷ 2.15L
Concentration of solution = 0.93M
Therefore, the concentration of a solution of 192 grams dissolved to make 2150 mL of solution is 0.93M.
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Answer:
it is phosphorus it has o most 2 times as much electromagnetically
Explanation: