<span>The </span>abundance of a chemical element<span> is a measure of the </span>occurrence<span> of the </span>element<span> relative to all other elements in a given environment. Abundance is measured in one of three ways: by the </span>mass-fraction<span> (the same as weight fraction); by the </span>mole-fraction<span> (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the </span>volume-fraction<span>. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and </span>ideal gas<span> mixtures. Most abundance values in this article are given as mass-fractions.
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Answer:
Explanation:
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In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:
Because we have four phosphorous and ten oxygen atoms at each side.
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Explanation:
Sulfur has 6 valence electron
Valence electron shows group( in column)
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
