Is it possible for the entropy of both a closed system and its surroundings to decrease during the process?

1 answer:

6 0

Only when heat is transferred from the system to its surroundings does a closed system suffer a decrease in entropy.

Only when heat is transferred from the system to its surroundings does a closed system suffer a decrease in entropy. Every internally reversible operation in a closed system generates entropy. Entropy remains constant in an adiabatic and internally reversible process of a closed system. Isolated systems' entropy cannot diminish.

When a system is not isolated but is in contact with its surroundings, the entropy of the open system may drop, requiring a balancing rise in the entropy of the surroundings. During a process, the entropy of an isolated system constantly increases, or in the case of a reversible process, remains constant (it never decreases).

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I’m pretty sure it would be D. Frying an egg

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4 years ago

Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou

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39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

$Ca(s) + Cl_2(g)$ → $CaCl_2(s)$

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

$m_{ca_C_l_2}$ = 16.7 g Ca x $\frac{1 mol \;of \;Ca}{40g Ca}$ x $\frac{1 mol \;of \;CaCl_2}{1 mol \;Ca}$ x $\frac{111g of \;CaCl_2}{1 mol \;CaCl_2}$

$m_{ca_C_l_2}$ = 39.96 g

Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

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you find that 7.36g of a compound has decomposed to give 6.93g of oxygen. the only other element in the compound is hydrogen. if

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First solve the moles of oxgen present in the compound

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then solve the moles of hydrogen present
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so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
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