Answer:
If 51.8 of Pb is reacting, it will require 4.00 g of O2
If 51.8 g of PbO is formed, it will require 3.47 g of O2.
Explanation:
Equation of the reaction:
2 Pb + O2 → 2 PbO
From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO
Molar mass of Pb = 207 g
Molar mass of O2 = 32 g
Molar mass of PbO = 207 + 32 = 239 g
Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO
= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO
Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.
If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2
Answer:
The velocity of the river increased.
There was more erosion in the stream.
The type of sediment that moved changed.
Explanation:
Answer:
7.98 × 10^3grams.
Explanation:
To find the mass of fluorine in the number of atoms provided, we first divide the number of atoms by Avagadros number (6.02 × 10^23atoms) to get the number of moles in the fluorine atom. That is;
number of moles (n) = number of atoms (nA) ÷ 6.02 × 10^23 atoms
n = 2.542 × 10^26 ÷ 6.02 × 10^23
n = 0.42 × 10^ (26-23)
n = 0.42 × 10^3
n = 4.2 × 10^2moles
Using mole = mass ÷ molar mass
Molar/atomic mass of fluorine (F) = 19g/mol
mass = molar mass × mole
Mass (g) = 19 × 4.2 × 10^2
Mass = 79.8 × 10^2
Mass = 7.98 × 10^3grams.
It would be 23, s choice C.
Answer:
n₂ =1.4 mol
Explanation:
Given data:
Mass of nitrogen = 2 g
Initial Volume occupy by nitrogen = 1.25 L
Final volume occupy by nitrogen = 25.0 L
Final number of moles = ?
Solution;
Formula:
V₁ / n₁ = V₂ / n₂
Number of moles of nitrogen:
Number of moles = mass/ molar mass
Number of moles = 2 g/ 28 g/mol
Number of moles = 0.07 mol
Now we will put the values in formula:
V₁ / n₁ = V₂ / n₂
n₂ = V₂× n₁ /V₁
n₂ = 25 L × 0.07 mol / 1.25 L
n₂ = 1.75 L. mol / 1.25 L
n₂ =1.4 mol
