Answer:
Atoms He (Avogadro’s number) → Moles of He (molar mass of He) → Mass of He
• molar mass of He (from the periodic table) = 4.003 g/mol
• Avogadro’s Number: Avogadro’s number gives us the number of entities present in 1 mole: 6.022 × 1023 He atoms in 1 mole of He
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Answer:
8.33 atm
Explanation:
Xe is 5 out of (4+5) or 5 / 9 ths of the gas present
5/9 * 15 atm = 8.33 atm
Answer:
So the volume will be 2.33 L
Explanation:
The reaction for the combustion is:
2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)
mass of butane to moles (mass / molar mass)
1.4 g / 58 g/mol
= 0.024 moles
2 moles of butane can produce 8 moles of carbon dioxide
0.024 moles of butane must produce (0.024 × 8) /2
= 0.096 moles of CO₂
Now we apply the Ideal Gases Law to find out the volume formed.
P . V = n . R . T
p = 1atm
n = 0.096 mol
R = 0.082 L.atm/mol.K
T = 273 + 23 = 296K
V = ?
1atm × V = 0.096 mol × 0.082 L.atm/mol.K × 296K
V = 0.096 mol × 0.082 L.atm/mol.K × 296K / 1atm
= 2.33 L
So the volume will be 2.33 L
Answer:
This element is Rubidium (Rb) and has an average atomic mass of 85.468 u
Explanation:
The average mass of an element is calculated by taking the average of the atomic masses of its stable isotopes.
The enitre atomic mass = 100 % or 1
⇒ this consists of X-85 with 72.17 % abundance with atomic massof 84.9118 g/mol
72.17 % = 0.7217
⇒ this consists of X-87 with 27.83 % abundance with atomic mass of 86.9092 g/mol
27.83 % = 0.2783
To calculate the mass of this isotope we use the following:
0.7271 * 84.9118 + 0.2783 * 86.9092 =85.468 g/mol
This element is Rubidium(Rb) and has an average atomic mass of 85.468 u
Answer:
Iodine
Explanation:
It's in the same group as chlorine.
