Answer:
That is NOT true!
Explanation:
No matter how little or big amount of waste you add, it is still pollution, and it adds to the big amount that is already there.
The mass for of aluminum that is produced by the decomposition of 5.0 Kg Al2O3 is 2647 g or 2.647 Kg
calculation
Write the equation for decomposition of Al2O3
Al2O3 = 2Al + 3 O2
find the moles of Al2O3 = mass/molar mass
convert 5 Kg to g = 5 x1000 = 5000 grams
molar mass of Al2O3 = 27 x2 + 16 x3 = 102 g/mol
moles =5000 g/ 102 g/mol = 49.0196 moles
by use of mole ratio between Al2O3 to Al which is 1:2 the moles of Al = 49.0196 x2 =98.0392 moles
mass of Al = moles x molar mass
= 98.0392 moles x 27g/mol = 2647 grams or 2647/1000 = 2.647 Kg
Answer:
0.6378 M
Explanation:
Molarity is defined by Moles per liter.
Plugging the given information in, we get (14.968 moles)/(23.47 L) which comes out to be about 0.6378 M
Answer:
4g/mol
Explanation:
Firstly, we can get the number of moles of the gas present using the ideal gas equation.
PV = nRT
Here:
P = 886 torr
V = 224ml = 224/1000 = 0.224L
T = 55 degrees celcius= 55+ 273.15 = 328.15K
R = molar gas constant = 62.36 L⋅Torr⋅K−1⋅mol−1
n = PV/RT
n = (886 * 0.224)/(62.36 * 328.15)
n = 0.009698469964 mole
Now to get the molar mass, this is mathematically equal to the mass divided by the number of moles. We have the mass and the number of moles, remaining only the molar mass.
First, we convert the mass to g and that is 38.8/1000 = 0.0388
The molar mass is thus 0.0388/0.009698469964 = 4g/mol
We use the gas law named Charle's law for the calculation of the second temperature. The law states that,
V₁T₂ = V₂T₁
Substituting the known values,
(0.456 L)(65 + 273.15) = (3.4 L)(T₁)
T₁ = 45.33 K
