To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
The correct answer is B) Basic. Hope this helps.
Answer:
86 mL
Explanation:
First find the moles of Pb (NO3)2
n=cv
where
c ( concentration)= 0.210 M
v ( volume in L) =0.05
n= 0.210 × 0.05
n= 0.0105
Using the mole ratio, we can find the moles of KCl by multiplying by 2
n (KCl) =0.0105 ×2
=0.021
v (KCl)= n/ c
= 0.021/ 0.244
=0.08606557377
=0.086 L
= 86 mL
Here, we have to get the number of atoms present in the 100 plane of the FCC crystal lattice.
There will be 2 atoms in 100 plane of FCC crystal lattice.
In the face centered crystal (FCC) lattice there are atoms at each corner of the cube and each are shared by 4 another atoms. And an atom is present at the face of the crystal.
For the 100 plane of the Miller indices the intercepts are a, ∞, ∞ or 2a, ∞, ∞.
Thus, for the 4 atoms of the corner at the cube shared by 4 other atoms will contribute, 4 ×
= 1 and the un-shared atoms at the face will contribute another 1, which make the total atom 1 + 1 = 2.
Answer:- 6984 kJ of heat is produced.
Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.
Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:

= 6984 kJ
So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.