Answer:
<em>D. One negative charge</em>
Explanation:
During the formation of a bond, if an atom gains an electron, after that it will be left with a negative charge compared to the atom before the bond is formed. This is because in these types of bonds, which are <em>ionic bonds</em>, there is a <em>transfer of electrons between atoms</em>, there will be one or more atoms that yield electrons that will be captured by another and other atoms that gain them, and the difference of charges produced by this transfer of electrons, will cause the union to occur due to the attraction between electrostatic forces.
If you have a neutral atom before joining, and it gains an electron to form a bond,<em> it will have one electron more than its initial state</em> (in the initial state, the number of protons and electrons is the same, because the atoms they are electrically neutral), so having an extra electron will make it have a negative charge, since there will be a difference between the number of protons and electrons that the atom possesses. <em>This is why the correct answer is D.
</em>
In the case of <em>response A and B</em>, <em>the atom could only remain positively charged if it loses electrons</em>, but as in this case it wins, <em>they are not correct</em>.
<em>The answer C is also not correct</em> because only one electron wins, so that it is left with two negative charges, <em>it should gain two electrons during the bond formation.</em>
Answer: Finding the [H3O+] and pH of Strong and Weak Acid Solutions The larger the Ka, the stronger the acid and the higher the H+ concentration at equilibrium. hydronium ion, H3O+, 1.0, 0.00, H2O, 1.0×10−14, 14.00.
Explanation:The hydrogen ion in aqueous solution is no more than a proton, a bare ... the interaction between H+ and H2O .
The answer is Thermal Energy :)
Answer:
The main advantage would be that with the pouring temperature being much higher, there is very little chance that the metal will solidify in the mould while busy pouring. This will allow for moulds that are quite intricate to still be fully filled. The drawbacks, though, include an increased chance defects forming which relates to shrinkage (cold shots, shrinkage pores, etc). Another drawback includes entrained air being present, due to the viscosity of the metal being low because of the high pouring temperature.
Na = 23 x 2.40 = 55.2
O = 16 x 2.40 = 38.4
H = 1 x 2.40 = 2.40
55.2 + 38.4 + 2.4 = 96
2.40 mol of NaOH = 96 amu
