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
Answer:
20.2 amu.
Explanation:
Let A represent isotope ²⁰X
Let B represent isotope ²²X
From the question given above, the following data were obtained:
For Isotope A (²⁰X):
Mass of A = 20
Abundance (A%) = 90%
For Isotope B (²²X):
Mass of B = 22
Abundance (A%) = 10%
Relative atomic mass (RAM) =?
The relative atomic mass (RAM) of the element can be obtained as follow:
RAM = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]
RAM = [(20 × 90)/100] + [(22 × 10)/100]
RAM = 18 + 2.2
RAM = 20.2 amu
Thus, relative atomic mass (RAM) of the element is 20.2 amu
Answer:As temperature increases, reactions take place. Generally, higher temperatures mean faster reaction rates; as molecules move about more quickly, reactant molecules are more likely to interact, forming products.
Explanation:
