Answer:
they had killed other animals
Answer:
XCH₄ = 0.461
XCO₂ = 0.539
Explanation:
Step 1: Given data
- Partial pressure of methane (pCH₄): 431 mmHg
- Partial pressure of carbon dioxide (pCO₂): 504 mmHg
Step 2: Calculate the total pressure in the container
We will sum both partial pressures.
P = pCH₄ + pCO₂
P = 431 mmHg + 504 mmHg = 935 mmHg
Step 3: Calculate the mole fraction of each gas
We will use the following expression.
Xi = pi / P
XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461
XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539
Answer:
For the first oxide, 1 g gives 0.888 g of copper.
Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.
For the second oxide, 1 g gives 0.798 g of copper.
Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.
So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.
Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.
Answer:
It's false ok it's non electrolyte
