Answer:
About 16.1 grams of oxygen gas.
Explanation:
The reaction between magnesium and oxygen can be described by the equation:

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.
Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:
- Convert grams of MgO to moles of MgO.
- Moles of MgO to moles of O₂
- And moles of O₂ to grams of O₂.
The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.
Dimensional analysis:

In conclusion, about 16.1 grams of oxygen gas was reacted.
You will obtain the same result if you compute with the 24.4 grams of Mg instead:

<h2>
Answer:
atoms</h2>
Explanation:
The given formula of the compound is 
The formula says
Every mole of
contains
moles of atoms of hydrogen.
Given that number of moles of compound is 
So,the number of moles of hydrogen atoms present is 
Since each mole has
atoms,
moles has
atoms.
Answer:
Cm = Cn/n = 0.1/2 = 0.05 (M)
n = 0.05*2= 0.1 mol
m H2SO4= 0.1*98= 9.8 g
m dd H2SO4= 9.8/ 0.98 =10 g
V= 10/ 1.84 = 5.435 ml
Explanation:
Answer:
a. 2.44 nm
b. 15.33 nm²
Explanation:
a. Calculate the thickness (in nm) of the surface film
Since the volume of a teaspoon equals V = 4.93 cm³ and the area of the oil film is half an acre = 1/2 × 4046.86 m² = 2023.43 m²
The volume of the oil film, V' equals the volume of the oil in the teaspoon, V
V' = Ah where A = cross-sectional area of oil film = 2023.43 m² = 2023.43 × 10⁴ cm² and h = height of oil film
So, V' = V
Ah = V
h = V/A
= 4.93 cm³/2023.43 × 10⁴ cm²
= 0.00244 × 10⁻⁴ cm
= 2.44 × 10⁻⁷ cm
= 2.44 × 10⁻⁷ cm × 1m/100 cm
= 2.44 × 10⁻⁹ m
= 2.44 nm
b. The surface area (in nm2) occupied by an oleic acid molecule on water.
Since the height of the oil film equals the diameter of the oil molecule, and the molecule is assumed to be a sphere of radius, r. Its surface area is thus A = 4πr²
r = h/2 = 2.44 nm/2 = 1.22 nm
A = 4πr²
A = 4π(1.22 nm)²
A = 4.88π nm²
A = 15.33 nm²
Answer:
The Forces that form each star would be your answer :)