Answer:
Group 1 or akali metals have the greatest metallic property.
Group 17 has the lowest metallic character.
C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.
Explanation:
Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.
Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.
Standard equation would be N2(g)+3H2(g)==>2NH3(g), so through stoichiometry, (4 mol N2)(2mol NH3/1 mol N2), assuming excess H2, would yield 8 moles of NH3.
Answer:
4.78 %.
Explanation:
<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>
<em></em>
<em>mass % = (mass of solute/mass of solution) x 100.</em>
<em></em>
mass of MgSO₄ = 50.0 g,
mass of water = d.V = (0.997 g/mL)(1000.0 mL) = 997.0 g.
mass of the solution = mass of water + mass of MgSO₄ = 997.0 g + 50.0 g = 1047.0 g.
<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (50.0 g/1047.0 g) x 100 = <em>4.776 % ≅ 4.78 %.</em>
Answer:
2.0 grams per cubic centimeter
Explanation:
(You can refer to the DMV triangle to help you solve this!)
Density = Mass/Volume
Density = 4.0g/2.0cm3
Density = 2.0g/cm3
Hope this helps!!!
-Unicorns110504
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Answer:
Cost to supply enough vanillin is 
Explanation:
Threshold limit of vanillin in air is
per litre means there should be
of vanillin in 1L of air to detect aroma of vanillin.

So, 
So amount of vanillin should be present to detect = 
As cost of 50 g vanillin is
therefore cost of
vanillin = 