Answer:
104.969 amu.
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 107.977 amu
Abundance (A%) = 0.1620%
Isotope B:
Mass of B = 106.976 amu
Abundance (B%) = 1.568%
Isotope C:
Mass of C = 105.974 amu
Abundance (C%) = 47.14%
Isotope D:
Mass of D = 103.973 amu
Abundance (D%) = 51.13%
Average atomic mass =?
The average atomic mass of the element can be obtained as follow:
Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]
Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]
= 0.175 + 1.677 + 49.956 + 53.161
= 104.969 amu
Therefore, the average atomic mass of the element is 104.969 amu.
<h3>
Answer:</h3>
Empirical formula is CrO
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of sample of Chromium as 7.337 gram
- Mass of the metal oxide formed as 9.595 g
We are required to determine the empirical formula of the metal oxide.
<h3>Step 1 ; Determine the mass of oxygen used </h3>
Mass of oxygen = Mass of the metal oxide - mass of the metal
= 9.595 g - 7.337 g
= 2.258 g
<h3>Step 2: Determine the moles of chromium and oxygen</h3>
Moles of chromium metal
Molar mass of chromium = 51.996 g/mol
Moles of Chromium = 7.337 g ÷ 51.996 g/mol
= 0.141 moles
Moles of oxygen
Molar mass of oxygen = 16.0 g/mol
Moles of Oxygen = 2.258 g ÷ 16.0 g/mol
= 0.141 moles
<h3>Step 3: Determine the simplest mole number ratio of Chromium to Oxygen</h3>
Mole ratio of Chromium to Oxygen
Cr : O
0.141 mol : 0.141 mol
1 : 1
Empirical formula is the simplest whole number ratio of elements in a compound.
Thus the empirical formula of the metal oxide is CrO
Answer:
=759.95 grams.
Explanation:
The molar mass of chromium is 51.9961 g/mol
Therefore the number of moles of chromium in 156 grams is:
Number of moles =mass/RAM
=156g/51.9961g/mol
=3 moles.
From the equation provided, 3 moles of chromium metal produce 2 moles of Chromium oxide.
Therefore 3 moles of chromium produce:
(3×2)/4 moles =1.5 moles of chromium oxide.
I mole of chromium oxide has a mass of 151.99 g
Thus 1.5 moles= 1.5mole ×151.99 g/mol
=759.95 grams.
Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Answer: YOU MAY FIND THE ANSWER IF YOU SEARCH THE WEB.
Explanation:
