Answer:
the weakest acid is B) HIO
Explanation:
pKa = - log Ka
the higher the value of pKa, the lower the dissociation, therefore, an acid will be stronger the lower its pKa.
a) HC2H3O2; Ka = 1.8 E-5
⇒ pKa1 = - Log (1.8 E-5) = 4.745
b) HIO; Ka = 23 E-11
⇒ pKa2 = - Log ( 23 E-11 ) = 9.638
c) HBrO; Ka = 23 E-9
⇒ pKa3 = - Log ( 23 E-9 ) = 7.638
d) HClO; Ka = 2.9 E-8
⇒ pKa4 = - Log ( 2.9 E-8 ) = 7.537
e) HCO2H; Ka = 63 E-5
⇒ pKa5 = - Log ( 63 E-5 ) = 3.200
from the values pKa, we places the acids from the weakest to the least weak:
1) pKa2; HIO (weakest)
2) pKa3
3) pKa4
4) pKa1
5) pKa5
Answer:

Explanation:
When it comes to electron configuration and orbitals, it's important to first identify what exactly we are trying to identify. Below is a given example:





Looking at the periodic table, identify the alkali metal family on the periodic table, or group one elements:

Notice how each configuration has an exponent of
, representative of a single electron in their s-orbital.
X4O10
Let molar mass of X be y
molar mass = 4y + 10 x 16 = 4y+160
so, moles = 85.2 / (4y+160)
Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0
so, 48 = 16 x 10 x [85.2 / (4y+160) ]
Solve the equation to get y.
y = 31
Answer:
joules
Explanation:
it is the measurement of energy
Answer:
- Question 19: the three are molecular compounds.
Explanation:
<em>Question 19.</em>
All of them are the combination of two kinds of different atoms in fixed proportions.
- C₂H₄: two carbon atoms per four hydrogen atoms
- HF: one hydrogen atom per one fluorine atom
- H₂O₂: two hydrogen atoms per two oxygent atoms
Thus, they all meet the definition of compund: a pure substance formed by two or more different elements with a definite composition.
Molecular compounds are formed by covalent bonds and ionic compounds are formed by ionic bonds.
Two non-metal elements, like H-F, C - C, C - H, H-O, H - H, and O - O will share electrons forming covalent bonds to complete their valence shell. Thus, the three compounds are molecular and not ionic.
<em>Question 20. </em>Formula of copper(II) sulfate hydrate with 36.0% water.
Copper(II) sulfate is CuSO₄. Its molar mass is 159.609g/mol
Water is H₂O. Its molar mass is 18.015g/mol
Calling x the number of water molecules in the hydrate, the percentage of water is:

From which we can solve for x:

Thus, there are 5 molecules of water per each unit of CuSO₄, and the formula is: