Answer is: <span>the size of the sample of uranium-238 does not affect its half-life.
</span><span>
The half-life for the
radioactive decay of U-238 is 4.5 billion years and is independent of initial
concentration (size of the sample).
</span>Half-life <span>is
the time required for a quantity (in this example number
of radioactive uranium) to reduce to half its initial
value.</span><span>
</span>
It is C. because there is a positive and a negative end of the battery. electron flow is simply from negative to positive.
Answer:
a push
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Explanation:
Answer:
3.925 mol.
Explanation:
- From the balanced equation:
<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g)
,</em>
It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂
.
<em>Using cross multiplication:</em>
4 moles of NaOH produced with → 1 mole of O₂
.
15.7 moles of NaOH produced with → ??? mole of O₂
.
<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) = <em>3.925 mol.</em>
The question is incomplete, here is the complete question:
Consider the following reaction: 
In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?
<u>Answer:</u> The average rate of appearance of oxygen gas is 
<u>Explanation:</u>
We are given:
Moles of oxygen gas = 
Volume of solution = 0.480 L
Molarity is calculated by using the equation:
So, 
The given chemical reaction follows:
The average rate of the reaction for appearance of 
is given as:
![\text{Average rate of appearance of }O_2=\frac{\Delta [O_2]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20rate%20of%20appearance%20of%20%7DO_2%3D%5Cfrac%7B%5CDelta%20%5BO_2%5D%7D%7B%5CDelta%20t%7D)
Or,
where,
= final concentration of oxygen gas = 0.0396 M
= initial concentration of oxygen gas = 0 M
= final time = 15.0 s
= initial time = 0 s
Putting values in above equation, we get:
Hence, the average rate of appearance of oxygen gas is
