Answer:
Land covers refer to all the manufactured structures and vegetation that covers the land it includes all vegetation including plants, shrubs, trees, and other man-made structures too.
On other hand, Land use is the term that explains the use of the land by the different human activities that are occurred on land that are directly related to the land.
Land cover influences land use by utilizing particular land such as parks, ponds, or other uses according to the land. and humans can cause changes in both when they urbanize the area or land.
Answer:7.229 grams of oxygen is formed by the complete reaction of 35.23 g of metallic sodium with oxygen at 130–200 °C, a process that generates sodium oxide, which in a separate stage absorbs oxygen: 4 Na + O2 → 2 Na2O. The ozone oxidizes the sodium to form sodium peroxide.
As given:
Initial moles of P taken = 2 mol
the products are R and Q
at equilibrium the moles of
R = x
total moles = 2 + x/2
Let us check for each reaction
A) P <-> 2Q+R
Here if x moles of P gets decomposed it will give 2x moles of Q and x moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = 2x
moles of R = x
Total moles = (2-x) + 2x + x = 2 +2x
B) 2P <-> 2Q+R
Here x moles of P will give x moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x
moles of R = x/2
Total moles = (2-x) + x + x/2 = 2 + x/2
C) 2P <-> Q+R
Here x moles of P will give x/2 moles of Q and x/2 moles of R
So at equilibrium
moles of P left = 2- x
moles of Q = x /2
moles of R = x/2
Total moles = (2-x) + x + x = 2
D) 2P <-> Q+2R
Here x moles of P will give x/2 moles of Q and x moles of R
So at equilibrium
moles of P left = 2-x
moles of Q = x/2
moles of R = x
Total moles = (2-x) + x/2 + x = 2 + x/2
Answer:
4.79 g of water
Explanation:
From the reaction equation;
C2H6(g) + 7/2O2(g) ----> 2CO2(g) + 3H2O(g)
Next we convert the given masses of reactants to moles of reactants.
For ethane; number of moles = mass/molar mass= 7.82g/ 30gmol-1= 0.261 moles
For oxygen; number of moles= 9.9 g/32gmol-1= 0.31 moles
Next we determine the limiting reactant, the limiting reactant yields the least amount of product.
For ethane;
From the reaction equation,
1 mole of ethane yields 3 moles of water
0.261 moles of Ethan yields 0.261 ×3 = 0.783 moles of water
For oxygen;
3.5 moles of oxygen yields 3 moles of water
0.31 moles of oxygen yields 0.31 × 3/3.5 = 0.266 moles of water
Hence oxygen is the limiting reactant.
Mass of water produced = 0.266 moles of water × 18gmol-1 = 4.79 g of water
