Answer: a) 
b) 
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
a) Mass of Ba= 66.06 g
Mass of Cl = 34.0 g
Step 1 : convert given masses into moles.
Moles of Ba =
Moles of Cl = \frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{34g}{35.5g/mole}=0.96moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Ba = 
For O =
The ratio of Ba: Cl= 1:2
Hence the empirical formula is
b) Mass of Bi= 80.38 g
Mass of O= 18.46 g
Mass of H = 1.16 g
Step 1 : convert given masses into moles.
Moles of Bi =
Moles of O= 
Moles of H=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Bi= 
For O =
For H=
The ratio of Bi: O: H= 1:3: 3
Hence the empirical formula is
Answer:
1) potential energy of the bond.
2) Linear
3) The electrons are transferred from K to Cl.
4) ClF
5) Oxygen
6) Electrolysis
7) Double displacement
Explanation:
As two atoms approach each other in a bonding situation, the potential energy of the bond is minimized as the internuclear distance of the bonding atoms decreases.
BeH2 has two electron domains and the central beryllium atom is sp2 hybridized. According to valence shell electron pair repulsion theory. A molecule having two regions of electron density will lead to a linear molecule.
KCl is an ionic compound hence there is a transfer of electrons from K(metal) to Cl(nonmetal).
ClF has partial charges because it contains a polar covalent bond. The partial charges arise from the dipole within the molecule. LiF is a pure ionic compound formed by transfer of electrons from Li to F. The species possess full and not partial charges.
When an oxygen atom bonds with another oxygen atom, what has been formed is a homonuclear covalent bond. Since the electro negativity of the both atoms is exactly the same, a pure covalent bond is formed. Recall that polar covalent bonds are formed when there is a significant electro negativity difference between the bonding atoms.
When direct current is passed through certain salt solutions during electrolysis, gases may be evolved and collected at the appropriate electrodes.
A double-replacement reaction is a reaction in which the cations and anions present in two different ionic compounds that are reacting together exchange their positions to form two new compounds on the product side. For instance, look at the reaction shown in question 7 as a typical example of this;
AgNO3 (s) + NaCl (s) → AgCl (s) + NaNO3 (s).
Answer:
293.15 K.
Explanation:
It is given that, the room temperature is 20 degrees Celsius.
We need to convert this temperature into kelvin.
The conversion from degrees Celsius to Kelvin is as follows :
We have, 
So,
So, the room temperature is 293.15 kelvin.
The answer is developer .
Chloe wants to lighten the color of her dark hair, which will require the use of both hair color and hydrogen peroxide.
Hydrogen peroxide (H₂O₂) is an example of a most commonly used developer in hair color.
Hydrogen peroxide lighten hair color, as it can dissolve darker pigment. According to the natural hair color and the length of time a person will leave hydrogen peroxide on hair, hair can turn somewhere on the color spectrum between caramel, orange and yellow.
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
