Sodium is a soft, silvery white, highly reactive element and is a member of the alkali metals within "group 1" (formerly known as ‘group IA’).
Potassium is a soft silvery-white metallic alkali metal that occurs naturally bound to other elements in seawater and many minerals
2007-09-30 21:36:56
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answer #1
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answered by Anonymous
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Absolutely! Sodium and Potassium (Na and K) are Alkali Earth metals in Group 1 of the periodic table. They are among some of the most reactive metals.
2007-09-30 21:02:51
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answer #2
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answered by Anonymous
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Yes they are. As defined, a metal is any of a category of electropositive elements that usually have a shiny surface, generally a good conductor of heat and electricity, and can be melted or fused, hammered into thin sheets, or drawn into wires. Typical metals form salts with nonmetals, basic oxides with oxygen, and alloys with one another metal.
It changes in form when combined with other elements so that it loses some or most of its metallic property.
2007-09-30 21:16:52
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answer #3
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answered by utsoy 1
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The silver nitrate (AgNO?) answer will finally turn blue. single replace reactions wherein a steel replaces yet another steel in simple terms artwork whilst the steel doing the changing is extra reactive than the steel being replaced. Out of the metals lead (Pb), silver (Ag), zinc (Zn), and cadmium (Cd), in simple terms silver (Ag) is way less reactive than the copper. hence, the only replace reaction 2AgNO? (aq) + Cu (s) ? Cu(NO?)? (aq) + 2Ag (s) works. The blue shade comes from the copper(II) nitrate that's formed.
2016-10-20 11:03:05
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answer #4
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answered by courts 4
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They are metals! no doubt at all about them. Not used for construction like steel, aluminium, copper etc., but very much metallic. They are alkali metals.
2007-09-30 21:22:38
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answer #5
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answered by Swamy 7
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They are Metals,
Their strong electropositive nature proves it
2007-09-30 21:04:49
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answer #6
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answered by rohit s BOSS... 2
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Yes, they are metals.
2007-09-30 21:01:26
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answer #7
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answered by ILoveMyDS 4
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yeppo
2007-09-30 21:17:35
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answer #8
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answered by Anonymous
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