1. The reactivity among the alkali metals increases as you go down the group due to the decrease in the effective nuclear charge from the increased shielding by the greater number of electrons. The greater the atomic number, the weaker the hold on the valence electron the nucleus has, and the more easily the element can lose the electron. Conversely, the lower the atomic number, the greater pull the nucleus has on the valence electron, and the less readily would the element be able to lose the electron (relatively speaking). Thus, in the first set comprising group I elements, sodium (Na) would be the least likely to lose its valence electron (and, for that matter, its core electrons).
2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.
3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.
4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
Answer:
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
Explanation:
Step 1: Data given
Initial temperature = 10.0 °C
Final temperature = 25.0 °C
Energy required = 30000 J
Mass of the object = 40.0 grams
Step 2: Calculate the specific heat capacity of the object
Q = m* c * ΔT
⇒With Q = the heat required = 30000 J
⇒with m = the mass of the object = 40.0 grams
⇒with c = the specific heat capacity of the object = TO BE DETERMINED
⇒with ΔT = The change in temperature = T2 - T2 = 25.0 °C - 10.0°C = 15.0 °C
30000 J = 40.0 g * c * 15.0 °C
c = 30000 J / (40.0 g * 15.0 °C)
c = 50 J/g°C
The specific heat capacity of the object is 50 J/g°C ( option 4 is correct)
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Answer:</h3>
C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)
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Explanation:</h3>
The balanced chemical equation for the combustion of the hydrocarbon in question is;
C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)
- A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
- Reactant side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
- Product side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
- An equation is balanced by putting appropriate coefficients on reactants and products involved in the reaction.
- An equation is balanced so as to obey the law of conservation of mass.
The best way to determine the number of atoms of arsenic in the sample will be to multiply 2.3 by Avagadro's number.
This is because Avagadro's number is the number of particles one mole of any substance has, and its value is 6.02 x 10²³
If the number of moles of a substance are known, then multiplying by Avagadro's number will give the number of particles. In this case, this is 1.38 x 10²⁴.
Answer:
Answer is explained below;
Explanation:
In 1904, after the discovery of the electron, the English physicist Sir J.J. Thomson proposed the plum pudding model of an atom. In this model, the atom had a positively-charged space with negatively charged electrons embedded inside it i.e., like a pudding (positively charged space) with plums (electrons) inside.
In 1911, another physicist Ernest Rutherford proposed another model known as the Rutherford model or planetary model of the atom that describes the structure of atoms. In this model, the small and dense atom has a positively charged core called the nucleus. Also, he proposed that just like the planets revolving around the Sun, the negatively charged electrons are moving around the nucleus.
By conducting a gold foil experiment, Rutherford disproved Thomson's model. In this experiment, positively charged alpha particles emitted from a radioactive source enclosed within a protective lead were used which was then focused into a narrow beam. It was then passed through a slit in front of which a thin section of gold foil was placed. A fluorescent screen (coated with zinc sulfide) was also placed in front of the slit to detect alpha particles which on striking the fluorescent screen would produce scintillation (a burst of light) which was visible through a microscope attached to the back of the screen.
He observed that most of the alpha particles passed straight through the gold foil without any resistance and this implied that atoms contain a large amount of open space. The slight deflection of some of the alpha particles, the large-angle scattering of other alpha particles and even the bouncing back of a very few alpha particles toward the source suggested their interactions with other positively charged particles inside the atom.
So, he concluded that only a dense and positively charged particle such as the nucleus would be responsible for such strong repulsion. Also, the negatively charged electrons electrically balanced the positive nuclear charge and they moved around the nucleus in circular orbits. Between the electrons and nucleus, there was an electrostatic force of attraction just like the gravitational force of attraction between the sun and the revolving planets.
Later, the Rutherford model was replaced by the Bohr atomic model.
