The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
The mass of water formed is
<u><em>calculation</em></u>
Use the ideal gas equation to calculate the moles of NH3 and O2
that is Pv= n RT
where; P= pressure,
V= volume,
n = number of moles,
R=gas constant = 0.0821 l .atm/ mol.K
make n the formula of the subject by diving both side by RT
n = PV /RT
The moles of NH3
n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles
The moles of O2
=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles
write the reaction between NH3 and O2
4 NH3 + 5 O2 →4 No +6H2O
from equation above 0.766 moles of NH3 reacted to produce
0.766 x 6/4 =1.149 moles of H2O
0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20
since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles
mass of H2O = moles x molar mass
from periodic table the molar mass of H2O = (1 x2)+16= 18 g/mol
mass = 18 g/mol x 0.9408 moles= 16.93 grams
46 is the answer. because if you add 26 and 20 that is the mass
Answer:
both spheres have a positive charge
ANSWER
A standard light microscope is used to view living organisms with little contrast to distinguish them from the background, which would be harder to see with the electron microscope.
Electron microscopes can be used to examine not just whole cells, but also the subcellular structures and compartments within them.
