-Positively charged nucleus
-Empty spaced
-Dense core
Answer:
2 H⁺ + 2e = H₂ ( reduction )
Explanation:
Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂
Fe( s ) = Fe⁺² + 2e ( oxidation )
2 H⁺ + 2e = H₂ ( reduction )
Answer:
A. there is an isotope of lanthanum with an atomic mass of 138.9
Explanation:
By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.
Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.
Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.
Thus, the correct option is:
<u>A. there is an isotope of lanthanum with an atomic mass of 138.9.</u>
Answer:
An ion with 11 protons, 11 neutrons, and 10 electrons would have a charge of 1+, also expressed as a charge of positive one or +1.
Answer:
a) equilibrium shifts towards the right
b) equilibrium shifts towards the right
c) equilibrium shifts towards the left
d) has no effect on equilibrium position
e) has no effect on equilibrium position
Explanation:
A reversible reaction may attain equilibrium in a closed system. A chemical system is said to be in a state of dynamic equilibrium when the rate of forward reaction is equal to the rate of reaction.
According to Le Chateliers principle, when a constraint such as a change in temperature, pressure, volume or concentration is imposed upon a system in equilibrium, the equilibrium position shifts in such a way as to annul the constraint.
When the concentration of reactants is increased, the equilibrium position is shifted towards the right hand side and more products are formed. For an endothermic reaction, the reverse reaction is favoured by a decrease in temperature. Increase in pressure has no effect on the system since there are equal volumes on both sides of the reaction equation. Similarly, the addition of a catalyst has no effect on the equilibrium position since it speeds up both the forward and reverse reactions to the same extent.
