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Ede4ka [16]
2 years ago
10

If the strength of the magnetic field at A is 10 units, the strength of the magnetic field at B is _____. 640 units 6.4 units 2.

5 units 1.25 units NEXT
Chemistry
2 answers:
MrMuchimi2 years ago
5 0

Answer:

2.5 units

Explanation:

dezoksy [38]2 years ago
4 0
Answer : Option C) 2.5 units

Explanation : For calculating the strength of a unknown magnetic field in comparison with the given magnetic field we can use the magnetic formula as;

\frac{ S_{1}}{S_{2}} =  \frac{ (D_{1}^{2}) }{(D_{2}^{2})} .

On substituting the values we get; 

\frac{10}{x}  =  \frac{4^{2}}{2^{2}} ;

On solving we get the magnetic B field strength as 2.5 units


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Chemical ____ are used to represent compounds ​
cupoosta [38]

Answer:

Its formula :)

Explanation:

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3 0
3 years ago
Read 2 more answers
A stock solution of HNO3 is prepared and found to contain 13.5 M of HNO3. If 25.0 mL of the stock solution is diluted to a final
salantis [7]

Answer : The correct option is, (C) 0.675 M

Explanation :

Using neutralization law,

M_1V_1=M_2V_2

where,

M_1 = concentration of HNO_3 = 13.5 M

M_2 = concentration of diluted solution = ?

V_1 = volume of HNO_3 = 25.0 ml  = 0.0250 L

conversion used : (1 L = 1000 mL)

V_2 = volume of diluted solution = 0.500 L

Now put all the given values in the above law, we get the concentration of the diluted solution.

13.5M\times 0.0250L=M_2\times 0.500L

M_2=0.675M

Therefore, the concentration of the diluted solution is 0.675 M

8 0
3 years ago
at a given tempature gas with a pressure of 150 kpa has volume of 0.8 l if the pressure decreases to 75kpa and the temperature r
vova2212 [387]
  <span>P*V/T=constant 
so P*V= constant*T 
if T doesn't change then 
P*V= constant 
so 150kPa*0.8L=75kPa*xL 
xL=150kPa*0.8L/75kPa=1.6L 
hope it help</span>
8 0
3 years ago
What is the density of a 12-gram sample of zinc with volume of 8.4 cm3? <br><br> pls hurry pls
Sindrei [870]

AnswerIm telling your teacher "Ms.Byrd" your in 8th grade and go to berry middle?

Explanation:

8 0
2 years ago
The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:
Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of AgNO_{3} in 50.0 mL of 0.100 M of AgNO_{3}

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= \frac{0.100}{1000}\times 50.0 moles

= 0.00500 moles

According to balanced equation, 1 mol of AgNO_{3} reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of AgNO_{3} react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = (100.0\times 1.00) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= \frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}

= \frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}

= -67716 J/mol

[m = mass, c = specific heat capacity, \Delta T = change in temperature and negative sign is included as it is an exothermic reaction]

4 0
3 years ago
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