Answer:
sulfur
Explanation:
sulfur has 4p electrons.
phosphorus has 3p electrons.
The question is asking which one has 4
Answer:
128g
Explanation:
Given parameters:
Mass of carbon = 48g
Mass of carbon dioxide = 176g
Unknown:
Mass of oxygen that reacted = ?
Solution:
Every chemical reaction must obey the law of conservation of mass. It states that "in a chemical reaction, matter is neither created nor destroyed" .
So;
Mass of carbon + Mass of oxygen = Mass of carbon dioxide
Mass of oxygen = Mass of carbon dioxide - Mass of carbon
Mass of oxygen = 176 - 48 = 128g
Answer:
All objects in motion do posses kinetic energy.
Explanation:
Answer:
Explanation:
An atom is the smallest unit of an element that can take part in a chemical reaction. Atoms are made up of protons, neutrons and electrons. Atoms can exist as a monoatomic (such as in the case of Helium, Xenon and Neon) or as diatomic (such as in the case of oxygen and nitrogen). Atoms take part in a chemical reaction and there reactivity varies among themselves.
From the above, it can be deduced that atoms have protons, neutrons and electrons. The number of protons (which is positively charged) of an atom determines it's position on the periodic table because elements in the periodic table are arranged according to the number of protons (called atomic number). The electron(s) present in the outermost shell of each atom (called valence electrons) determines there chemical reactivity. What happens here is that, all atoms (except noble gases) want to achieve there duplet or octet configuration so as to become stable. This octet configuration means they want to have there outermost shell completely filled (with eight electrons or two electrons for duplet). They usually achieve this configuration by taking part in chemical reactions. Thus, when an atom has just one electron in it's outermost shell, it becomes easy to lose it to another atom by way of interacting with it in a chemical reaction. When it loses this single electron (valence electron) in it's outermost shell, it becomes stable with the inner completely filled shell (that would be the new outermost shell). Examples include Lithium, sodium and potassium. Sodium (with eleven electrons and three shells) would lose the single electron in it's outermost shell so as to have just two shells with the second shell completely filled with eight electrons. Thus, <u>the more the valence electron to be lost to achieve the octet structure</u>,<u> the lesser the reactivity of the atom</u>.
Also, an atom that has just one electron to complete it's own outermost shell and thus achieve it's octet structure is also highly reactive. This is also because it is easy for this atom to receive a single electron and become completely filled. Examples include chlorine, fluorine and iodine. Fluorine (with nine electrons and two shells) will easily accept one more electron so as to achieve it's octet structure with a completely filled outermost shell (of eight electrons). Thus, <u>the lesser the electrons to be gained to achieve the octet configuration, the higher the chemical reactivity of such atoms</u>. Noble gases have extremely low or no reactivity at all for this reason because it has a completely filled outermost shell (no losing or donating).
It should also be noted that metals (which are found on the left of the periodic table) exist as monoatomic while gases (which are found on the right), with the exception of noble gases, are mostly diatomic.
Answer:
(i). C6H2COOH and Na2CO3(aq)
observation: <u>Bubbles</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>colourless</u><u> </u><u>gas</u><u> </u><u>(</u><u>carbon</u><u> </u><u>dioxide</u><u> </u><u>gas</u><u>)</u>
(ii) CH3CH2CH2OH and KMnO4 /H
observation: <u>The</u><u> </u><u>orange</u><u> </u><u>solution</u><u> </u><u>turns</u><u> </u><u>green</u><u>.</u>
[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>oxidation</em><em> </em><em>of</em><em> </em><em>propanol</em><em> </em><em>to</em><em> </em><em>propanoic</em><em> </em><em>acid</em><em> </em><em>occurs</em>]
(iii) CH3CH2OH and CH3COOH + conc. H2SO4
observation: <u>A</u><u> </u><u>sweet</u><u> </u><u>fruity</u><u> </u><u>smell</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>
[<em>This</em><em> </em><em>is</em><em> </em><em>because</em><em> </em><em>an</em><em> </em><em>ester</em><em>,</em><em> </em><em>diethylether</em><em> </em><em>is</em><em> </em><em>formed</em><em>]</em>
(iv) CH3CH = CHCH3 and Br2 /H2O
observation: <u>a</u><u> </u><u>brown</u><u> </u><u>solution</u><u> </u><u>is</u><u> </u><u>formed</u><u>.</u>