1 and 5 are definitely significant. The zeros might not be significant .
Answer:
After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.
Explanation:
<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
Answer:
have you tried c
Explanation:
the chicken and I don't know if you can make it
A solution has a pOH of 7. 1 at 10∘c. Then the pH of the solution given that kw=2. 93×10−15 at this temperature is 7.4 .
It is given that,
pOH of solution = 7.1
Kw =2.93×10^(-15)
Firstly, we will calculate the value of pKw
The expression which we used to calculate the pKw is,
pKw=-log [Kw]
Now by putting the value of Kw in this expression,
pKw =−log{2.93×10^(-15)}
pKw =15log(2.93)
pKw=14.5
Now we have to calculate the pH of the solution.
As we know that,
pH+pOH=pKw
Now put all the given values in this formula,
pH+7.1=14.5
pH=7.4
Therefore, we find the value of pH of the solution is, 7.4.
learn more about pH value:
brainly.com/question/12942138
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