Answers:
8.70 g
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s <em>gather all the information</em> in one place.
M_r: 32.00 44.01
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 18H₂O
m/g: 9.88
(a) Calculate the <em>moles of O₂
</em>
n = 9.88 g O₂ ×1 mol O₂ /32.00 g O₂
n = 0.3088 mol O₂
(b) Calculate the <em>moles of CO₂</em>
The molar ratio is (16 mol CO₂/25 mol O₂)
n = 0.3088 mol O₂ × (16 mol CO₂/25 mol O₂)
n = 0.1976 mol CO₂
(c) Calculate the <em>mass of CO₂
</em>
Mass of CO₂ = 0.1976 mol CO₂ × (44.01 g CO₂/1 mol CO₂)
Mass of CO₂ = 8.70 g CO₂
Chemical formula of the glucose: C₆H₁₂O₆
We calculate the molar mass:
atomic mass (C)=12 u
atomic mass (H)=1 u
atomic mass (O)=16 u
atomic weight (C₆H₁₂O₆)=6(12 u)+12(1u)+6(16 u)=72 u+12u+96 u=180 u.
Therefore : 1 mol of glucose will be 180 g
The molar mass would be: 180 g/ mol
2) we calculate the number of moles of 1.5 g.
180 g---------------------1 mol
1.5 g---------------------- x
x=(1.5 g * 1 mol) / 180 g≈8.33*10⁻³ moles
we knows that:
1 mol = 6.022 * 10²³ particles (atoms or molecules)
3)We calculate the number of molecules:
Therefore:
1 mol-----------------------6.022*10²³ molecules of glucose
8.33*10⁻³ moles-------- x
x=(8.33*10⁻³ moles * 6.022*10²³ molecules)/1 mol≈5.0183*10²¹ molecules.
4)We calculate the number of C, H and O atoms:
A molecule of glucose have 6 atoms of C, 12 atoms of H, and 6 atoms of O,
number of atoms of C=(6 atoms/1 molecule)(5.0183*10²¹molecules)≈
3.011*10²²
number of atoms of H=(12 atoms/1 molecule)(5.0183*10²¹ molecules)≈
6.022*10²² .
number of atoms of O=(6 atoms/1 molecule)(5.0183*10²¹ molecules)≈
3.011*10²²
Answer: we have 3.011*10²² atoms of C, 6.022*10²² atoms of H, and 3.011*10²² atoms of O.
Answer:
Air & Water
Explanation:
Air and water is the common mixtures in the book of the book called "Science & Land"
Answer:
B.They do not react chemically
Explanation:
This is because all noble gases haves full outer shell therefore they don’t participate in bonding.They are referred to as inert which means unreactive.
