The reactants are C6H12O6 (Glucose) and 6O2 (Oxygen) and the products are 6CO2 (Carbon Dioxide) and 6H20 (Water)
The Lewis bases are those which can donate electron pair to an acceptor. These electrons can be a pair of electron that is a lone pair, a negative charge or electrons in a 
-bond.
- In
, the presence of -bond make it a Lewis base. - In
, the presence of negative charge makes it a Lewis base. - In
, the presence of lone pair on nitrogen makes it a Lewis base. - In
, there is no lone pair, no -bond, no negative charge so it is not a Lewis base.
Hence, , , and are Lewis bases.
Answer:
1.306 × 10^24 formula units.
Explanation:
The chemical equation of this qkrstios as follows:
Fe + 2H2O → FeO2 + 2H2
Based on the reaction, 2 mole of H2O produces 1 mole of FeO2
Using the formula as follows;
Mole = mass/molar mass
Molar mass of H2O = 1(2) + 16 = 18g/mol
mole = 78/18
mole = 4.33moles
If 2 mole of H2O reacts to produce 1 mole of FeO2
4.33 moles of H2O will produce 4.33/2 = 2.17moles of FeO2.
To convert moles to formula units, we multiply number of moles by Avagadro number (6.02 × 10^23 units)
= 2.17 moles × 6.02 × 10^23 units
= 13.06 × 10^23
= 1.306 × 10^24 formula units.
Answer:
B im so sorry if its wrong im sure it right
Explanation:
see it has the lowest specific heat so it wouldnt cool because of the bionestint particles
Answer:
C. 30.0 g NaOH and add water until the final solution has a volume of 3.00 L.
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity
V = volume of solution in liter ,
n = moles of solute ,
from the question ,
M = 0.250M
V = 3.00 L
M = n / V
n = M * v
n = 0.250M * 3.00 L = 0.75 mol
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
n = 0.75 mol NaOH
m = molecular mass of NaOH = 40 g/mol
n = w / m
w = n * m
w = 0.75 mol * 40 g/mol = 30.0 g
Hence , by using 30.0 g of NaOH and dissolving it to make up the volume to 3 L , a solution of 0.250 M can be prepared .
