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Mama L [17]
4 years ago
13

In this problem, you will use Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire

loop.
You will need to use the right-hand rule to find the direction of the induced current. When the switch is open, which of the following statements about the magnetic flux through the wire loop is true? Assume that the direction of the vector area of the wire loop is to the right.
A) There is no magnetic flux through the wire loop.
B) There is a positive flux through the wire loop.
C) There is a negative flux through the wire loop.
Chemistry
1 answer:
Vsevolod [243]4 years ago
4 0

Answer:

The correct answer is Option A (There is no magnetic flux through the wire loop.)

Explanation:

Magnetic flux measures the entire magnetic field that passes through the wire loop.

The right hand rule can be demonstrated on how magnetic flux is generated through the moving current in the wire loop. The magnetic flux through the wire loop  will decrease as it moves upward through the magnetic field region.

If the direction of the vector area of the wire loop is to the right, and the switch is closed, it will push the magnetic flux to the right which will now be increased due to an equal increase in the current in the wire loop. But, when the switch is open, this will halt the movement of current through the wire loop thus affecting the generation of magnetic field. This would make the magnetic flux to be zero.

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If the number of moles of a gas initially contained in a 2.10 l vessel is doubled, what is the final volume of the gas in liters
Anon25 [30]
Avagadro's law gives the relationship between volume of gas and number of moles of the gas. When the temperature and pressure are constant, volume is directly proportional to number of moles of gas.
\frac{V1}{n1} =  \frac{V2}{n2}
V -volume and n - number of moles of gas.
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
initial number of moles n1 - N
n2 - 2N as the number of moles are doubled 
V1 - initial volume - 2.10 L
V2 - final volume - V
substituting the values in the equation 
\frac{2.10L}{N}  \frac{V}{2N}
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7 0
3 years ago
What is the rate law for the reaction 2A + 2B + 2C --> products
-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

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How much Ca(NO3)2 should be weighed out to have 0.650 mol?A) 66.4 g.B) 97.68.C) 107 g.D) 133 g.E) 165 g.
Bogdan [553]

Answer:

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Explanation:

Step 1: Calculate the molar mass of Ca(NO₃)₂

We can calculate the molar mass of Ca(NO₃)₂ by adding the masses of its elements.

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M = 1 × 40.08 g/mol + 2 × 14.01 g/mol + 6 × 16.00 g/mol

M = 164.10 g/mol

Step 2: Calculate the mass corresponding to 0.650 moles of Ca(NO₃)₂

We multiply the number of moles by the molar mass.

0.650 mol × 164.10 g/mol = 107 g

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