There are 6 protons, 7 neutrons, and 6 electrons since the atomic mass is rounded to 13.
Answer:
there are approximately n ≈ 10²² moles
Explanation:
Since the radius of the earth is approximately R=6378 km= 6.378*10⁶ m , then the surface S of the earth would be
S= 4*π*R²
since the water covers 75% of the Earth's surface , the surface covered by water Sw is
Sw=0.75*S
the volume for a surface Sw and a depth D= 3 km = 3000 m ( approximating the volume through a rectangular shape) is
V=Sw*D
the mass of water under a volume V , assuming a density ρ= 1000 kg/m³ is
m=ρ*V
the number of moles n of water ( molecular weight M= 18 g/mole = 1.8*10⁻² kg/mole ) for a mass m is
n = m/M
then
n = m/M = ρ*V/M = ρ*Sw*D/M = 0.75*ρ*S*D/M = 3/4*ρ*4*π*R² *D/M = 3*π*ρ*R² *D/M
n=3*π*ρ*R² *D/M
replacing values
n=3*π*ρ*R² *D/M = 3*π*1000 kg/m³*(6.378*10⁶ m)² *3000 m /(1.8*10⁻² kg/mole) = 3*π*6.378*3/1.8 * 10²⁰ = 100.18 * 10²⁰ ≈ 10²² moles
n ≈ 10²² moles
The density of the stone is 5 because the formula is mass/volume and the volume is 5 and the mass is 25
Answer:a formula giving the number of atoms of each of the elements present in one molecule of a specific compound.
Explanation:
Answer:
21.8 grams.
Explanation:
Molar mass data from a modern periodic table:
How many moles of MgO will be produced if Mg is the limiting reactant?
Number of moles of Mg:
.
The ratio between the coefficient of Mg and that of MgO is 2:2. Two moles of Mg will make two moles of MgO. 0.670644 moles of MgO will be produced if Mg is the limiting reactant.
How many moles of MgO will be produced if O₂ is the limiting reactant?
Number of moles of O₂:
.
The ratio between the coefficient of O₂ and that of MgO is 1:2. One mole of O₂ will make two moles of MgO.
of MgO will be produced if O₂ is in excess.
How many moles of MgO will be produced?
0.541284 is smaller than 0.670644. Only 0.541284 moles of MgO will be produced since O₂ will run out before all 16.3 grams of Mg is consumed.
What's the mass of 0.541284 moles of MgO?
Formula mass of MgO:
.
Mass of 0.541284 moles of MgO:
.