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ahrayia [7]
2 years ago
9

When a phasor-domain circuit has dependent sources, you should not use a sequence of source transforms to find the thévenin and

norton equivalents for that circuit, and you cannot use the equivalent impedance technique to find the equivalent impedance. instead, you must find the open-circuit phasor voltage (which is the thévenin voltage) and the short-circuit phasor current (which is the norton current). then you can use ohm's law with the phasor voltage and current to find the thévenin (or norton) impedance. remember that you can also use the test source method to find the thévenin impedance. consider the circuit shown here. in this circuit, the phasor voltage source vs=10∠0∘ v. we wish to find the norton equivalent of this circuit to the left of terminals a and
b?
Chemistry
1 answer:
Ierofanga [76]2 years ago
4 0
Yes but the answer is A

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insens350 [35]
B is the answer or check it out
6 0
3 years ago
Please help This is for my chemistry class and I need to get it done but I’m lost and need answers
Vanyuwa [196]

Answer:

chlorine is a non metal.

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5 0
2 years ago
Determine the equilibrium constant, Keq, at 25°C for the reaction
adelina 88 [10]

Explanation:

The given chemical reaction is:

2Br^- (aq) + I_2(s)  Br_2(l) + 2I^- (aq)

E^ocell=oxidation potential of anode + reduction potential of cathode\\

The relation between Eo cell and Keq is shown below:

deltaG=-RTlnK_e_q\\delta=-nFE^o cell\\=>nFE^o cell=RTlnK_e_q\\lnK_e_q=\frac{nF}{RT} E^o cell

The value of Eo cell is:

Br- undergoes oxidation and I2 undergoes reduction.

Reduction takes place at cathode.

Oxidation takes place at anode.

Hence,

E^ocell= (-1.07+0.53)V\\=-0.54V

F=96485 C/mol

n=2 mol

R=8.314 J.K-1.mol-1

T=298K

Substitute all these values in the above formula:

ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3

Answer:

Keq=6.13x10^33

3 0
3 years ago
If 0.0106 g of a gas dissolves in 0.792 L of water at 0.321 atm, what quantity of this gas (in grams) will dissolve at 5.73 atm?
Alex73 [517]

Answer:

0.189 g.

Explanation:

  • This problem is an application on <em>Henry's law.</em>
  • Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
  • Solubility of the gas ∝ partial pressure
  • If we have different solubility at different pressures, we can express Henry's law as:

<em>S₁/P₁ = S₂/P₂,</em>

S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm

S₂ = ??? g/L and P₂ = 5.73 atm

  • So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.

<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>

<em></em>

8 0
3 years ago
How many milliliters of 1.50 M magnesium sulfate soulution is required to supply 2.50 mole of this salt?
arsen [322]

Answer:

1670 ml

Explanation:

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Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.

3 0
3 years ago
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