Answer:
the answer should 126 859.2 m2
The average the natural abundances of the various isotopes of carbon to arrive at the fractional mass.
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>
Reduction is the process by which eletrons are gained by an atom.
Answer: <span>The molecules of a substance which must have the
<u>a</u></span>
<u>bility to move past one another</u> are said to be flexible.
Explanation: Those substances are said to be flexible which can be
bent without breaking. There are many substances which are
hard in nature but still can be bent. The hardness of such materials is due to
strong interactions between the molecules and the flexibility comes due to their
amorphous backbone. Therefore, greater the
crystalline level of macromolecules lesser is the flexibility and greater the amorphous character greater is the flexibility and vice versa. Also, the flexibility of polymers is increased by adding
plastisizers in it. Plastisizers make the hard polymers flexible by breaking the crosslinkers and enabling the macromolecules to move past one another.