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den301095 [7]
3 years ago
11

A proton travels at a speed of 2.0 × 106 meters/second. Its velocity is at right angles with a magnetic field of strength 5.5 ×

10-3 tesla. What is the magnitude of the magnetic force on the proton?
Chemistry
1 answer:
Finger [1]3 years ago
6 0

So here we are given that the the velocity of the proton ( V ) is 2.0 × 10^6 meters / second, with a magnetic field of strength 5.5 × 10^{-3}  tesla. If they each form a right angle, they are hence perpendicular to one another, such that ....

F = q( V × B ),

F = q v B( sin ∅ ),

F = q v B( sin( 90 ) )

.... they form the following formula. Let's go through each of the variables in our formula here -

{ F = Magnetic Force ( which has to be calculated ), q = charge of proton (has charge of 1.602 × 10^{19} coulombs ), B = magnetic field }

All we have to do now is plug and chug,

F = ( 1.602 × 10^{19} )( 2.0 × 10^6 )( 5.5 × 10^{-3} ) = ( About ) 1.8 × 10^{-15} Newtons

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Answer:

See explanation

Explanation:

Full molecular equation;

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Full ionic equation

2NH3(aq) + Ag^+(aq) + NO3^-(aq) --------> [Ag(NH3)2]^+(aq) + NO3^-(aq)

Net ionic equation;

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3 years ago
Which is a characteristic of mixtures?
d1i1m1o1n [39]

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8 0
3 years ago
Which of the following are characteristics of a gas? (Select all that apply.)
ladessa [460]

Answer:

it has a fixed volume it can also be compressed

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4 0
2 years ago
Read 2 more answers
The teacher tells your group to make a stock solution of sodium chloride, and then diluting it to
sergejj [24]

The idea here is that you need to figure out how many moles of magnesium chloride,

MgCl

2

, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

c

=

n

V

So, how many moles of magnesium chloride must be present in the target solution?

c

=

n

V

⇒

n

=

c

⋅

V

n

=

0.158 M

⋅

250.0

⋅

10

−

3

L

=

0.0395 moles MgCl

2

Now determine what volume of the target solution would contain this many moles of magnesium chloride

c

=

n

V

⇒

V

=

n

c

V

=

0.0395

moles

3.15

moles

L

=

0.01254 L

Rounded to three sig figs and expressed in mililiters, the volume will be

V

=

12.5 mL

So, to prepare your target solution, use a

12.5-mL

sample of the stock solution and add enough water to make the volume of the total solution equal to

250.0 mL

.

This is equivalent to diluting the

12.5-mL

sample of the stock solution by a dilution factor of

20

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3 0
2 years ago
Why are non-metals dull in appearance as solids?
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3 years ago
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