Answer:
ummm according 2 me babies are consumers and rocks are also consumers
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.
1.
Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar
2. For the volume, let's find the total volume first.
V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
The answer to this question will be C
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:
16.6 g of Al are produced in the reaction of 82.4 g of AlCl₃
Explanation:
Let's see the decomposition reaction:
2AlCl₃ → 2Al + 3Cl₂
2 moles of aluminum chloride decompose to 2 moles of solid Al and 3 moles of chlorine gas.
We determine the moles of salt:
82.4 g . 1mol/ 133.34g = 0.618 moles
Ratio is 2:2. 2 moles of salt, can produce 2 moles of Al
Then, 0.618 moles of salt must produce 0.618 moles of Al.
Let's convert the moles to mass → 0.618 mol . 26.98g /mol = 16.6 g
