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Rashid [163]
2 years ago
12

EASY!!!!! ILL MARK BRAINLIEST!!!!!!!! why does measured current differ from calculated current

Chemistry
1 answer:
sergey [27]2 years ago
8 0

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:

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In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
If HCl is a 0.05 M solution and 50 mL is used to titrate the NaOH, what is the molarity of NaOH if the flask contains 100 mL?
Liono4ka [1.6K]

Answer:

Molarity of NaOH = 0.025 M

Explanation:

Given data:

Molarity of  HCl = C₁ = 0.05 M

Volume of HCl = V₁= 50 mL

Molarity of NaOH = C₂=?

Volume of NaOH =V₂= 100 mL

Solution:

Formula:

C₁V₁  = C₂V₂

C₁ = Molarity of  HCl

V₁  = Volume of HCl

C₂ = Molarity of NaOH

V₂ = Volume of NaOH

Now we will put the values:

C₁V₁  = C₂V₂

0.05 M × 50 mL = C₂ × 100 mL

2.5 M.mL =C₂ × 100 mL

C₂  = 2.5 M.mL /100 mL

C₂  = 0.025 M

7 0
3 years ago
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USPshnik [31]

Answer:

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<em />

<em>m = </em><u><em>25</em></u><em> </em><em> </em><u><em>Kg</em></u>

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