Zeff = Z - S
Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.
For boron, the electronic configuration is 1s₂ 2s₂ 2p₄
Z = 5, S = 2
Zeff = 5-2 = +3
For O, electronic configuration is 1s₂ 2s₂ 2p₄
Z = 8, S = 2
Zeff = 8-2 = +6
Hence, the correct answer is second option, that is, +3 and +6, the Zeff of boron is smaller in comparison to O, thus, boron exhibits a bigger size than O.
Answer:
D) contains more OH– ions than H+ ions
Explanation:
THESE ARE THE OPTIONS FOR THE QUESTION
A) causes some indicators to change color B) conducts electricity C) contains more H+ ions than OH– ions D) contains more OH– ions than H+ ions
Base can be regarded as substance which can dissociates in water then form hydroxide ions (OH–) .
Bases can be regarded as compounds which break into hydroxide ions I.e (OH-) with more other compounds if put in an aqueous solution. Hence it contains more OH– ions
<u>Given:</u>
Surface area at the narrow end, A1 = 5.00 cm2
Force applied at the narrow end, F1 = 81.0 N
Surface area at the wide end, A2 = 725 cm2
<u>To determine:</u>
Force F2 applied at the wide end
<u>Explanation:</u>
Use the relation
F1/A1 = F2/A2
F2 = F1*A2/A1 = 81.0 N * 725 cm2/5.00 cm2 = 11,745 N
Ans: (b)
The force applied at the wide end = 11,745 N
The melting point of pure lead : 327 °C
<h3>Further explanation</h3>
Given
Amount of tin and melting point of solder
Required
The melting point
Solution
The composition of solder = tin and lead
So if it is 100% tin, 0% lead or 0% tin, 100% lead
From the table it is shown that when the position is 100% tin in the solder, the melting point of the solder is 232 °C, so it shows that the melting point of pure lead is obtained when% tin in solder = 0 (100% lead in solder), so that the melting point is obtained. : 327 °C
We have to calculate the molar mass of AL(OH)₃
Atomic mass (Al)=27 amu
Atomic mass (O)=16 amu
Atomic mass (H)=1 amu
molecular weight= 27 amu+3(16 amu + 1 amu) =78 amu.
Therefore, the molar mass of Al (OH)₃ is 78 g/ mol
Now, we calculate the number of moles in 98.3 g of aluminum hydroxide.
78 g-------------------1 mol
98.3 g----------------- x
x=(98.3 g * 1 mol) / 78 g=1.26 moles.
Answer: 1.26 moles.
