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Answer:</h2>
<h2>Half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive ..!!</h2><h2 /><h2 />
<h2>Hopefully u will satisfy with my answer..!!</h2><h2>Please Mark on brainleast please..!!</h2>
The answer is 1.6 atm. Let's first calculate the mole fraction of gas Y.
The mole fraction (x) is: x = n1/n, where n1 is a number of moles of an
individual gas in a mixture and n is total moles of the gas mixture. We
know that n1 of gas Y is 6.0 mol (n1 = 6.0 mol) and that there are in
total 8. mol of the gas mixture (n = 2.0 + 6.0 = 8.0 mol). Now calculate
the mole fraction of gas Y. x = 6.0/8.0 = 0.75. Now, let's use the mole
fraction of gas Y (x) and the total pressure (P) to calculate the
partial pressure of gas Y (P1): x = P1/P. P1 = x * P. If x = 0.75 and P =
2.1 atm, then the partial pressure of gas Y is: P1 = 0.75 * 2.1 atm =
1.6 atm.
<h3>Answer:</h3>
3.2 moles of Oxygen
<h3>Explanation:</h3>
Acetic Acid having chemical formula C₂H₄O₂ and structural formula attached below is the second member of carboxylic family in organic compounds. It is commonly used as vinegar<em> i.e</em>. a mixture containing 5 % Acetic acid and 95 % water.
As shown in structure 1 mole of Acetic acid contains 2 moles of Carbon atoms, 4 moles of Hydrogen atoms and 2 moles of Oxygen atoms respectively. Hence, the number of moles of Oxygen atoms contained by acetic acid containing 3.2 moles of Carbon is calculated as,
2 moles of C accompany = 2 moles of O
So,
3.2 moles of C will accompany = X moles of O
Solving for X,
X = (3.2 mol C × 2 mol O) ÷ 2 mol C
X = 3.2 mol of Oxygen
The answer is .0004704. i hope that helped.
