Hot water holds less dissolved oxygen than cold water.
b. Negative ΔH
c. Positive ΔG
d. Negative ΔG
e. Positive ΔS
Answer:
The rubber becomes brittle and can break in your hand. The explanation for why this happens concerns cross-linking bonds. Ultra-violet light from the sun provides the polymer molecules with the activation energy they need to be able to form more cross-links with other chains.
If excessively excessive cross-links are formed in rubber, it becomes overly rigid and brittle, hampering its natural elasticity and strength. The increase in cross-links restricts the moving of the polymer chains, undermining the effectiveness of the rubber in many applications.
The formation of cross-links in rubber significantly affects its properties. In the case where too many cross-links are formed, the rubber is likely to become overly rigid and brittle. This is because the cross-links restrict the movement of the polymer chains, which reduces flexibility and elasticity. As a result, too many cross-links can compromise the usefulness of rubber for many applications, which require its natural elasticity and strength. For example, in a rubber band, if too many cross-links were formed, then it would be less stretchy and snap more easily when stretched.
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b) 22.7
c) 55.6
d) 31.8
e) 45.5
The amount of oxygen needed to oxidize 12.5g of propane is 45.5g.
Stoichiometry is the calculation of the quantitative relationships between the amounts of reactants and products in chemical reactions.
The balanced chemical equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
To calculate the amount of oxygen needed to react with 12.5g of propane, we need to use stoichiometry and the molar mass of propane. The molar mass of propane is 44.1 g/mol.
First, we need to convert the mass of propane to moles:
12.5 g C3H8 × (1 mol C3H8 / 44.1 g C3H8) = 0.283 mol C3H8
According to the balanced equation, 5 moles of oxygen are required to react with 1 mole of propane. So, we can calculate the amount of oxygen needed as follows:
0.283 mol C3H8 × (5 mol O2 / 1 mol C3H8) = 1.415 mol O2
Finally, we can convert the moles of oxygen to grams using the molar mass of oxygen, which is 32.0 g/mol:
1.415 mol O2 × 32.0 g O2/mol = 45.1 g O2
Therefore, the answer is e) 45.5 g of oxygen is needed to oxidize 12.5g of propane (C3H8).
Learn more about Stoichiometry here:
brainly.com/question/30215297?
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