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Viktor [21]
3 years ago
13

How to balance the equation NaNo3 + PbO & Pb(NO3)2 + Na2O

Physics
1 answer:
dedylja [7]3 years ago
6 0
It would be "Double replacement".

Hope this helps!
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The 25-lb slender rod has a length of 6 ft. using a collar of negligible mass, its end a is confined to move along the smooth ci
sp2606 [1]
Which of the following is NOT a factor in efficiency?<span><span>A.metabolism</span><span>B.type of movement</span><span>C.muscle efficiency</span><span>D.digestion</span></span>
3 0
3 years ago
Which is the transfer of thermal energy through the motion of particles caused by temperature differences? Question 20 options:
nevsk [136]
Answer is Conduction.
7 0
3 years ago
Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono
kari74 [83]

Answer:

Por lo tanto, la presión del ozono es:

P=0.011\: atm  

Explanation:

Podemos usar la ecuacion de los gases ideales;

PV=nRT (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol

Finalmente, usando la ecuacion 1 despejamos la presion P

P=\frac{nRT}{V}

P=\frac{0.095*0.082*324.09}{224}  

Por lo tanto, la presion del ozono es:

P=0.011\: atm  

Espero te haya ayudado!

5 0
3 years ago
you push a ball to star it rolling along a "perfectly frictionless" surface. How far will the ball roll
-BARSIC- [3]
Technically, it should roll forever.
6 0
3 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
4 years ago
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