To determine the shapes of molecules, we must become acquainted with the Lewis electron dot structure. Although the Lewis theory does not determine the shapes of molecules, it is the first step in predicting shapes of molecules. The Lewis structure helps us identify the bond pairs and the lone pairs.
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The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Gallum: Z = 31
electron configuration: [Ar] 4s^2 3d10 4s2 4p1
Highest energy electron: 4p1
Quantum numbers:
n = 4, because it is the shell number
l = 1, it corresponds to type p orbital
ml = may be -1, or 0, or +1, depending on space orientation, they correspond to px, py, pz
ms = may be -1/2 or +1/2, this is the spin number.
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
