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Tatiana [17]
3 years ago
15

An object of mass 5 kilograms is moving across a surface in a straight line with a speed of 3.5 meters/second. What amount of fo

rce would be required to keep the object in motion if you ignore the sureface friction?
Chemistry
1 answer:
vampirchik [111]3 years ago
4 0

F = MA A = change in velocity Constant velocity = no change in velocity = Zero acceleration Ignoring friction, 5 kg * 0 m/s^2 = 0 N Therefore, there is no force (0 N) required to keep the object in motion if the surface friction is ignored.

I hope my answer has come to your help.


Read more on Brainly.com - brainly.com/question/2005686#readmore

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What element has 28 electrons when electrically neutral
muminat

Answer:

nickel

It also tells you the number of electrons that the element has in its outside shells. If the atomic number of nickel is 28 then every atom of nickel has 28 protons in its nucleus and 28 electrons outside the nucleus.

Explanation:

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3 years ago
How many moles of N2O5 are needed to produce 7.90 g of NO2? 2N2O5 = 4NO2 + O2
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3 years ago
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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

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Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

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Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

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Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

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Answer:

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8 0
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Answer:

4

Explanation:

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hope this helps :)

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