To measure the distance between two opposite sides of an object
Answer:- 
Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.
We know that, 1 US dollar = 100 cents
1 cent = 1 US penny
So, 1 US dollar = 100 US pennies
Let's make the set up starting with 1 penny as:
=
Therefore, we can bye of Cf in one US penny.
Electrons in an atom can be classified as core electrons and valence electrons. Valence electrons are those electrons which are present in valence shell and participates in bond formation. While, Core electrons are all remaining electrons which are not present in valence shell, hence not take part in bonding.
Atomic number of Selenium (Se) is 34 hence it has 34 electrons with following electronic configuration;
1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
From electronic configuration it is found that the valence shell is 4, and the number of electrons present in valence shell are 6. So,
Core Electrons = Total Electrons - Valence Electrons
Core Electrons = 34 - 6
Core Electrons = 28
Result:
There are 28 core electrons in Selenium.
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+ 
⇄ 
Formation constant Kf
Kf = / ( [][] ) = 5.0 × 10¹⁰
Now,
[] = 
; ∝₄ = 0.35
so the equilibrium is;
+ 
⇄ + 4H⁺
Given that; 
= { 1 mol reacts with 1 mol }
so at equilibrium, = = x
∴
+ ⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [][] ) = / ( [][] ) = 5.0 × 10¹⁰
⇒ / ( [][] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
