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nikdorinn [45]
2 years ago
14

"Technician A says that Ohm's law can be used to determine circuit current flow if total circuit resistance and total voltage ar

e known. Technician B says that Ohm's law can be used to calculate the unknown resistance of a load in a circuit if total current and source voltage are known. Which technician is correct?
Computers and Technology
1 answer:
Troyanec [42]2 years ago
6 0

Answer:

Both Technicians are correct.

Explanation:

Remember that Ohm's Law its a relationship between Voltage (E), Current (I) and Resistance (R) in an electrical circuit. This relationship is defined by the following equation:

E=I*R

Each one of the letters could be called a variable. Now, also remember that the number of equations you have is equal to the number of variables unknown you could have.

In Ohm's Law you have 3 variables (E,I,R) and 1 equation. It means that you need to know 2 variables (Whatever 2 variables) to know the third.

Technician A says, in other words, that could calculate the Circuit current flow (I) knowing total circuit resistance (R) and total voltage (E). Note that Technician A knows 2 to 3 variables, then it's possible to use Ohm's Law with the following equation:

I=\frac{E}{R}

Technician B says, in other words, that could determine the unknown resistance (R) knowing total current (I) and source voltage (E). Once again, note that Technician B knows 2 to 3 variables, then it's possible to use Ohm's Law with the following equation:

R=\frac{E}{I}

It is concluded that both technicians are right.

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2 years ago
What is the rate constant at 25.0 ∘c based on the data collected for trial b?
Alecsey [184]
<h2><u>Answer: </u></h2>

acetone + I2 + HCl ---> iodated acetone  

Equation:

rate = k * [acetone]^x * [I2]^y * [HCl]^z  

Once we know x, y, z, we can plug in any of the trials A->D and determine k  NOTE: We can't use run E because temperature has an effect on rate. E was run at a different temperature.

The first thing to note is that do NOT have concentrations. We have volumes at a given molarity.  

Table:  

.001M I2.. ..050M HCl.. .1.0M acetone.. .water.. temp..time.. total vol  

.. ....mL.. ... ... .. .mL.. .. ... .. .. mL.. .. .. .. ..mL.. ..°C.. .sec.. .. .. .L  

A.. ...5.. .. .. .. .. ..10.. .. .. .. .. ..10.. .. .. .. ..25.. .. .25.. .130.. .. 0.05  

B.. ..10.. .. .. ... .. 10.. .. .. ... .. .10.. .. .. ... ..20.. . .25.... 249.. ..0.05  

C.. . 10.. .. .. .. .. .20.. .. .. .... .. 10.. .. ... ... .10... ..25.. ..128... .0.05  

D.. . 10.. ... .. ... ..10.. .. ... .. ... 20.. .. .. .. .. 10.. .. 25.. ..131.. ..0.05  

E.. ..10.. ... .. ... ..10.. ... ... ... ..10.. ... .. ... .20.. .. 42.8.. .38.. ..0.05  

We can translate that into molarity in solution using this formula. (molarity pure ingredient * mL used / 1000 / total volume in liters)  

 

.. .. ..I2.. .... HCl.. acetone.. temp.. ..rxn time  

.. .. ..M.. .. ...M.. .... .M.. .... ..°C.. .. .. sec  

A.. 0.0001.. 0.01..... 0.2.. .... .25... .... .130  

B.. 0.0002.. 0.01.. .. 0.2.. .. .. 25.. .. .. .249  

C.. 0.0002.. 0.02.. .. 0.2.. .. .. 25.. .. .. .128  

D.. 0.0002.. 0.01.. .. 0.4.. .. . .25.. ... ...131  

E.. 0.0002.. 0.01.. .. 0.2.. . ....42.8.. .. .. 38  

From runs B and D, we can see that rate dropped by half .

As [I2] and [HCl] were held constant and [acetone] was doubled.  

This means x=-1 in this equation  

Rate = k * [acetone]⁻¹ * [I2]^y * [HCl]^z  

Rate = k * [I2]^y * [HCl]^z  

From runs A and B  

[I2] doubles  

[HCl] remains the same  

[acetone] remains the same  

rate doubles   as [I2] doubles, rate doubles  

y = 1   rate = k * [acetone]⁻¹ * [I2]¹ * [HCl]^z  

And from runs B and C, we can see that  , As [HC] doubles, (all else equal) the rate halves.

Z = -1  

Rate = k * [I2] / ([acetone] * [HCl])  

Rearranging  

k = rate * [acetone] * [HCl] / [I2]  

From any experimental run (A-D), we can calculate k.

using A to calc k... ..k = 2600 M²/sec  

using B to calc k... . k = 2490 M²/sec  

using C to calc k... . k = 2560 M²/sec  

using D to calc k... . k = 2620 M²/sec  

NOTE.. the problem statement said to use the data from run B to calc k.

Hence Final Answer:

k = 2490 M²/sec

4 0
3 years ago
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