Answer:
Hydrogen +
Explanation:
An acid is a chemical which "wants" to donate some protons, or hydrogen + ions. Since a hydrogen atom is just a proton and an electron, the ion lacking an electron is simply a proton. Hope this helps!
Answer:
± 1 or ± 2
Explanation:
Electrovalent bonds are chemical bonds that are established on the premise of transferring electrons between two atoms.
In this bond type, a higly electronegative atom, typically a non-metal receives electrons from an atom with lesser electronegativity, a metal.
To know the number of electrons involved in forming electrovalent bonds, we typically look at the groups of atoms that combines to form the bond.
Metals are found in group I and II on the periodic table. Metals are electropositive and are good electron donors. These metals have 1 and 2 electrons in their valence shell respectively. In like manners, the more electronegative atoms are found in group VI and VII. The elements in these groups are non-metals with high electronegativity and requires just 1 and 2 electrons to complete their octet.
10) In order to find the conjugate acid of a chemical you just add a hydrogen to the chemical.
examples: the conjugate acid of Cl⁻ is HCl, the conjugate acid of PO₄³⁻ is HPO₄²⁻, the conjugate acid of NH₃ is NH₄⁺, the conjugate acid of HCO₃⁻ is H₂CO₃, and the conjugate acid of H₂O is H₃O⁺
To find the conjugate base of a chemical you just reverse that process (take away a hydrogen).
examples: the conjugate base of H₂SO₄ is HSO₄⁻, the conjugate base of CH₃COOH is CH₃COO⁻, the conjugate base of H₃PO₄ is H₂PO₄⁻, and the conjugate base of H₂O is OH⁻.
When you identify conjugate acids and bases in a reaction you look to see what lost a proton and what gained a proton. The chemical that gave up the proton acted as an acid and produced a conjugate base while the chemical that accepted a proton produced a conjugate acid.
Example: HCl+NaOH⇒NaCl+H₂O The acid is HCl and its conjugate base is Cl⁻ while NaOH was the base and H₂O is the conjugate acid. (you can ignore the sodium since it is a spectator ion).
11) When completing acid base reactions, need to identify the acid and the base since the acid will give a proton the base creating a conjugate base of the acid and conjugate acid of the base. (You need to balance the equation after you determine what the products will be)
example: H₂SO₄+2NaOH⇒Na₂SO₄+2H₂O (SO₄²⁻ is the conjugate base of HSO₄⁻ which is the conjugate base of H₂SO₄. HSO⁻ is created with the first NaOH molecule and then SO₄⁻ is created with the second NaOH.)
12) All acid base reaction form a salt consisting of the cation from the base and anion from the acid.
examples: NaCl could have come from NaOH reacting with HCl. K₃PO₄ could have come from KOH and H₃PO₄.
13) I don't really know how you are supposed to solve it with out knowing the Ka value of H₂S. H₂S is a weak acid and therefore will not dissociate completely in water so the only way of being able to find the concentration of H⁺ ions that dissociate is knowing the Ka value of H₂S and using ice tables. (Ka=[H⁺][A⁻]/[HA] and is basically the equilibrium constant for the acid when put into water where A⁻ is the conjugate base and HA is the acid).
14) Ca(OH)₂ is a strong base and will therefore dissociate completely in water. That means that when you find the concentration of OH⁻ in solution you can multiply that by the volume of the solution (in liters) to find the number of moles of OH⁻. Then you can divide that by 2 to find the number of moles of Ca(OH)₂ needed. pOH=14-pH which means that pOH=4.2. [OH⁻]=10^-pOH which means [OH⁻]=6.3x10^-5 M. 6.3x10^-5Mx3.00L=1.89x10^-4mol OH⁻ which means that (1.89x10^-4)/2=9.46x10^-5mol Ca(OH)₂.
I hope this helps. Let me know if anything is unclear.
<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M
