Answer:
55,124.729438282
Explanation:
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Answer:
No component is perfect. All have tolerances that can vary. If you construct a simple circuit where a 10 volt power supply feeds a 10 ohm resistor, you would expect to measure a current of one ampere. BUT - the wiring has some resistance too. This adds perhaps 0.1 ohms to the circuit. The resistor has a +-5% tolerance. If it is 5% high, it may measure 10.5 ohms. That's a total circuit resistance of 10.6 ohms. The power supply may have a tolerance of +-1%. Suppose it's 1% low. That's an output of 9.9 volts in real life. So you have 9.9 volts dropped across 10.6 ohms. you will measure closer to 0.934 amps instead of 1.000 amps. To make matters worse, most electronic components have a temperature coefficient, that is, their values change with different temperatures. You may get a completely different reading tomorrow if the temperature is different! Finally, with current measurements in particular, you are inserting the ammeter in series with the circuit under test. Ammeters have some inherent resistance too, so by putting the ammeter in the circuit, you are changing the very current you are trying to measure (a little)! Oh yeah, the ammeter has a tolerance too. Its reading may be off a little even if everything else is perfect. Sometimes you have to wonder how we get a decent reading at all. Fortunately the errors are usually fairly small, and not all tolerances are off in the same direction or off the maximum amount. They tend to cancel each other out somewhat. BUT - in rare circumstances everything CAN happen like I said, and the error can be huge.
Explanation:
The way you can solve this, is by using this equation to solve for percent acidity:
%acidity = (grams of acetic acid / grams of vinegar) * 100.
Hope this helps!
Ancient cyanobacteria released hydrogen which assisted in creating the atomsphere as we know it today.
We are asked to provide the net ionic equation for the reaction of HF (aq) and NaF (aq). HF is a weak acid and is in the following equilibrium:
HF (aq) ⇄ H⁺ (aq) + F⁻ (aq)
Meanwhile, NaF (aq) is an ionic compound that will dissociate completely in aqueous solutions:
NaF (aq) → Na⁺ (aq) + F⁻ (aq)
We can combine the ionic species with HF, as we are told to show F⁻ as a reactant:
HF (aq) + Na⁺ (aq) + F⁻ (aq) → HF (aq) + Na⁺ (aq) + F⁻ (aq)
We can eliminate the spectator ions, which in this case are Na⁺ ions, and this leaves us with the net ionic equation involving F⁻:
HF (aq) + F⁻ (aq) → HF (aq) + F⁻ (aq)
In this instance, the proton is just transferred between F⁻ ions and the end result is the formation of more HF, so there is no net reaction taking place.