A colligative property is when the addition of a solute to a solvent alters the boiling point or The freezing point of a solvent
1 answer:
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Cu + 2H2SO4 ⟶ CuSO4 + SO2 + 2H20
In left hand side of the equation.
Cu = 1 atom
H = 4 atoms
S = 2 atoms
O = 8 atoms
In right hand side of the equation.
Cu = 1 atom
S = 2 atoms
0 = 8 atoms
H = 4 atoms
• All the atoms are balanced in the left and right side of the equation and it satisfies the law of conservation of mass.
• Equation is balanced and correct.
*See the attachment .
LiBr.
<h3>Explanation</h3>Note that the group number in this answer refers to the new IUPAC group number, which ranges from 1 to 18. Counts from the left. Start with the first two column (group 1 and 2), go on to the transition elements (Sc, Ti, etc. in group 3 through 12), and continue with the nonmetals (group 13 through 18).
Li is a group 1 metal. As a metal, it tends to form positive ions ("cations"). Metals in group 1 and 2 are <em>main group</em> metals. The charge on main group metal ions tends to be the same as the group number of the metal. Li is in group 1. The charge on an Li ion will be +1. Formula of the Li ion will be .
Br is a group 17 nonmetal. As a nonmetal, it tends to form negative ions ("anions"). The charge on nonmetal ions excepting for H tends to equal the group number of the nonmetal minus 18. Br is in group 17. The charge on a Br ion will be 17 - 18 = -1. Formula of the Br ion will be
All the ions in an ionic compound carry charge. However, some of the ions like are positive. Others ions like
are negative. Charge on the two types of ions balance each other. As a result, the compound is <em>overall</em> neutral.
1 × (+1) + 1 × (-1) = 0. The positive charge on one ion balances the negative charge on one
ion. The two ions would pair up at a 1:1 ratio.
The empirical formula for an ionic compound shows all the ions in the compound. Positive ions are written in front of negative ions. is positive and
is negative. The formula shall also show the simplest ratio between the ions. For the compound between Li and Br, a 1:1 ratio will be the simplest. The "1" subscript in an empirical formula can be omitted. Hence the formula: LiBr.
Answer:
pH = 9,57
[M²⁻] = 7,948x10⁻²M
[HM⁻] = 4x10⁻⁵M
[H₂M] = 0M
Explanation:
The moles of maleic acid presents in the solution are:
0,923g×=7,952x10⁻³moles of H₂M
60,0mL of 0,265M KOH are:
0,0600L×=0,0159 moles of KOH
The reactions of maleic acid (H₂M) and then with HM⁻ are:
H₂M + KOH → HM⁻ + H₂O + K⁺ (1)
HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)
For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:
0,0159 - 7,952x10⁻³ = <em>7,948x10⁻³ moles of KOH</em>
By (2), the moles produced of M²⁻ are the same as moles of KOH, <em>7,948x10⁻³ moles, </em>and moles of HM⁻ are:
7,952x10⁻³ -<em> </em>7,948x10⁻³ <em> = 4x10⁻⁶ moles of HM⁻</em>
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [M²⁻] /[HM⁻]
pH = 6,27 + log₁₀ <em>7,948x10⁻³ / 4x10⁻⁶</em>
pH = 9,57
The moles of M²⁻ are 7,948x10⁻³ and volume of the solution is 0,1000L,
[M²⁻] = 7,948x10⁻²M
Moles of HM⁻ are 4x10⁻⁶:
[HM⁻] = 4x10⁻⁵M
And there is not H₂M:
[H₂M] = 0M
I hope it helps!