Ans: The final volume of the balloon is 4.5 L
<u>Given:</u>
Volume of balloon inflated with 3 breaths = 1.7 L
<u>To determine:</u>
Volume of balloon after a total of 3+5 = 8 breaths
<u>Explanation:</u>
Volume of the balloon per breath = 1.7 L * 1 breath/3 breaths = 0.567 L
Final volume of balloon after 8 breaths = 0.567 L * 8 breath/1 breath
= 4.536 L
Answer:
The mass of xenon in the compound is 2.950 grams
Explanation:
Step 1: Data given
Mass of XeF4 = 4.658 grams
Molar mass of XeF4 = 207.28 g/mol
Step 2: Calculate moles of XeF4
Moles XeF4 = mass XeF4 / molar mass XeF4
Moles XeF4 = 4.658 grams / 207.28 g/mol
Moles XeF4 = 0.02247 moles
Step 3: Calculate moles of xenon
XeF4 → Xe + 4F-
For 1 mol xenon tetrafluoride, we have 1 mol of xenon
For 0.02247 moles XeF4 we have 0.02247 moles Xe
Step 4: Calculate mass of xenon
Mass xenon = moles xenon * molar mass xenon
Mass xenon = 0.02247 moles * 131.29 g/mol
Mass xenon = 2.950 grams
The mass of xenon in the compound is 2.950 grams
Answer:
Avogadro’s number was calculated by determining the number of atoms in 12.00 g of carbon-12.
Explanation:
The number of particles presents in one mole of a substance is known as Avogadro's number.
Avogadro's number is 
atoms or molecules or ions or particles present in one mole of a substance. It is denoted by the symbol 
or 
. It is a dimensionless quantity.
Avogadro's number was proposed by Jean Perrin but named in the honor of italian scientist Amedeo Avogadro.
Avogadro's number is the number of atoms present in 12 grams of carbon-12.
We are given with a compound, Methane (CH4), with a molar
mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in
g. We need to solve first the molecular weight of Methane, that is
C=12 g/mol
H=1g/mol
CH4= 12 g/mol +1(4) g/mol = 16 g/mol
With 0.893 mol sample, its corresponding mass is
g CH4= 0.893 mol x 16g/mol =14.288 g
Therefore, the mass of methane is 14.288 g
Answer:
Half life = 1600 years
Explanation:
Given data:
Total mass of sample = 45.00 g
Mass remain = 5.625 g
Time period = 4800 years
Half life of radium-226 = ?
Solution:
First of all we will calculate the number of half lives passes,
At time zero 45.00 g
At first half life = 45.00 g/ 2= 22.5 g
At 2nd half life = 22.5 g/ 2 = 11.25 g
At 3rd half life = 11.25 g/ 2= 5.625 g
Half life:
Half life = Time elapsed / number of half lives
Half life = 4800 years / 3
Half life = 1600 years
