Answer:
40 seconds chief just double it
Explanation:
3.0 × 10¹¹ RBC's (or) 3E11 RBC's
Solution:
Step 1: Convert mm³ into L;
As,
1 mm³ = 1.0 × 10⁻⁶ Liters
So,
0.1 mm³ = X Liters
Solving for X,
X = (0.1 mm³ × 1.0 × 10⁻⁶ Liters) ÷ 1 mm³
X = 1.0 × 10⁻⁷ Liters
Step 2: Calculate No. of RBC's in 5 Liter Blood:
As given
1.0 × 10⁻⁷ Liters Blood contains = 6000 RBC's
So,
5.0 Liters of Blood will contain = X RBC's
Solving for X,
X = (5.0 Liters × 6000 RBC's) ÷ 1.0 × 10⁻⁷ Liters
X = 3.0 × 10¹¹ RBC's
Or,
X = 3E11 RBC's
Answer : The mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
Explanation : Given,
Mass of oxygen in sulfur dioxide = 3.49 g
Mass of sulfur in sulfur dioxide = 3.50 g
Mass of oxygen in sulfur trioxide = 9.00 g
Mass of sulfur in sulfur trioxide = 6.00 g
Now we have to calculate the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide.
Mass of oxygen per gram of sulfur for sulfur dioxide = 
Mass of oxygen per gram of sulfur for sulfur dioxide = 
and,
Mass of oxygen per gram of sulfur for sulfur trioxide =
Mass of oxygen per gram of sulfur for sulfur trioxide = 
Thus, the mass of oxygen per gram of sulfur for sulfur dioxide and sulfur trioxide is, 0.997 g and 1.5 g respectively.
PH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.
or basically 2
Answer:
See explanation
Explanation:
In an atom, the inner electrons may shield the outer electrons from the attractive force of the nucleus. We, refer to this phenomenon as the <u><em>shielding effect</em></u>, It is defined as a decrease in the magnitude of attraction between an electron and the nucleus of an atom having more than one electron shell (energy level).
Shielding effect increases down the group due to addition of more shells but decreases across the period due to the increase in the size of the nuclear charge.
As the magnitude of shielding increases down the group, ionization of electrons becomes easier and the first ionization energies of elements decreases as we move down the group. Since shielding effect decreases across the period, the first ionization energies of elements increases across the period.
