Answer: The gas generated by two antacid tablets has a smaller volume.
Explanation:
Since the antiacid is the limiting reagent, we know that the more tablets there are, the more gas there will be.
This means that there will be more gas generated by the four antiacid tablets when compared to the two antiacid tablets, which gives us that the gas generated by the two antiacid tablets has a smaller volume.
Answer:
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Answer:
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure 2.2.2 ). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Explanation:
Covalent
bonds = sharing of electrons between two atoms of the same elements or elements
close to each other on the periodic table. Usually they are metals sometimes
non-metals. In polar bonds electrons are
shared unequally. Non polar bonds share electrons equally.
Answer:
1.64 moles O₂
Explanation:
Part A:
Remember 1 mole of particles = 6.02 x 10²³ particles
So, the question becomes, how many '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?
This implies a division of given number of particles by 6.02 x 10²³ particles/mole.
∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂
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Part B needs an equation (usually a combustion of a hydrocarbon).
