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Tpy6a [65]
3 years ago
9

A student designed an experiment to test the effect on the magnitude of the magnetic field that is generated around the wire loo

p. The student increased the current that passes through the wire and tests the magnitude of the magnetic force that is generated. The experimental setup is shown here. Which statements accurately describe this experimental design? Select ALL that apply. A) The student's dependent variable is the size of the wire loop. B) The student's independent variable is the size of the wire loop. C) The student's dependent variable is the magnitude of the magnetic field that is generated. D) The student's independent variable is the magnitude of the magnetic field that is generated. E) The student's independent variable is the amount of current that is being passed through the wire.
Chemistry
1 answer:
kykrilka [37]3 years ago
5 0

Answer: C ) the student’s dependent variable is the magnitude of the magnetic field that is generated

E ) the student’s independent variable is the amount of current that is being passed through the wire

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
katovenus [111]

Answer:

The new partial pressures after equilibrium is reestablished:

PCl_3,p_1'=6.798 Torr

Cl_2,p_2'=26.398 Torr

PCl_5,p_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=p_1=13.2 Torr

Partial pressure of the Cl_2=p_2=13.2 Torr

Partial pressure of the PCl_5=p_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{p_1}{p_1\times p_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=p_1'=13.2 Torr

Partial pressure of the Cl_2=p_2'=?

Partial pressure of the PCl_5=p_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=p_1'+p_2'+p_3'

263.0Torr=13.2 Torr+p_2'+217.0 Torr

p_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{p_3'}{p_1'\times p_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished:

p_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

p_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

p_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

6 0
3 years ago
A piece of metal that has a density of 5.2 g/cm3 and a mass of 100 g was placed in a full jar of water. How many mL of water was
Klio2033 [76]

Answer:

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Answer:

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The answer is B.

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