Answer:
this is the answer
I hope it helps you although
Given :
A student has a 1 g sample of each of the following compounds: NaCl, KBr, and KCl.
To Find :
The samples in order of increasing number of moles in the sample.
Solution :
Molecular mass of NaCl, KBr, and KCl is 58.5 g/mol , 119 g/mol and
74.5 g/mol respectively .
Moles of NaCl ,
.
Moles of KBr ,
.
Moles of KCl ,
.
The order of moles in increasing order is :
KBr , KCl and NaCl .
Hence , this is the required solution .
The final temperature of the copper is 59.0. The specific heat capacity of copper is 0.38 j/g -k
Answer:
The correct answer is B. Since the two metals have the same mass, but the specific heat capacity of iron is much greater than that of gold, the final temperature of the two metals will be closer to 498 K than to 298 K
Explanation:
Iron is hotter and gold is colder, therefore, according to laws of thermodynamics, iron will lose heat to gold until they are at the same temperature.
The specific heat capacity of iron(0.449) is over three times that of gold(0.128). Since masses are equal, this means that each time iron's temperature drops by one degree, the energy released it releases makes gold's temperature increase by more than 3 degrees. So gold's temperature will be climbing much faster than iron's is falling. Meaning they will meet closer to the initial temperature of iron than that of gold
Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.