Answer:
H₂ is excess reactant and O₂ the limiting reactant
Explanation:
Based on the chemical reaction:
2H₂(g) + O₂(g) → 2H₂O
<em>2 moles of H₂ react per mole of O₂</em>
<em />
To find limiting reactant we need to convert the mass of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
10g H₂ * (1mol / 2.016g) = 4.96 moles
<em>Moles O₂ -Molar mass: 32g/mol-:</em>
22g O₂ * (1mol / 32g) = 0.69 moles
For a complete reaction of 0.69 moles of O₂ are needed:
0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂
As there are 4.96 moles,
<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>
Answer:
11.9g remains after 48.2 days
Explanation:
All isotope decay follows the equation:
ln [A] = -kt + ln [A]₀
<em>Where [A] is actual amount of the isotope after time t, k is decay constant and [A]₀ the initial amount of the isotope</em>
We can find k from half-life as follows:
k = ln 2 / Half-Life
k = ln2 / 27.7 days
k = 0.025 days⁻¹
t = 48.2 days
[A] = ?
[A]₀ = 39.7mg
ln [A] = -0.025 days⁻¹*48.2 days + ln [39.7mg]
ln[A] = 2.476
[A] = 11.9g remains after 48.2 days
<em />
They are located in the U.S.A
2 in front of H2. Technically nothing in front of O2 otherwise put a 1 then a 2 in front of H2O
2H20 + 1O2 —— > 2H2O
The very common mineral shown in the figure that is referred in this problem that is commonly a pink- to cream-colored mineral with wavy, light-colored lines and does not effervesce would be feldspar. It make up about 41 percent weight of the Earth's crust. It is a group of rocks that contains tectosilicate compounds.
