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2 years ago

4. Distinguish where kinetic energy and potential energy are the greatest in the loop.

Chemistry

1 answer:

aksik [14]2 years ago

4 0

Kinetic energy is the energy from movement. Hence the fastest movement is at the bottom of the loop - therefore Kinetic energy is highest at the flat bottom of the loop.

Potential energy is the opposite - it occurs at the top of the loop when the car moves the slowest

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Someone help please ...

Tanzania [10]

Answer:

Yes

Explanation:

4 0

3 years ago

Why is it so difficult to experience the northern lights here in michigan? I need to know noww

Anni [7]

Answer:

Think about where you're located on the globe. That'll tell you your answer

Explanation:

7 0

3 years ago

1. Unas de las formas de producir nitrógeno gaseoso (N2) es mediante la oxidación de metilamina (CH3NH2), tal como se muestra en

Maslowich

Answer:

a) 4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

b) m = 5,043 g

c) % = 69,4 %

Explanation:

a) La ecuación balanceada es la siguiente:

4CH₃NH₂ + 9O₂ ⇄ 4CO₂ + 10H₂O + 2N₂

En el balanceo, se tiene en la relación estequiométrica que 4 moles de metilamina reacciona con 9 moles de oxígeno para producir 4 moles de dióxido de carbono, 10 moles de agua y 2 moles de nitrógeno.

b) Para determinar la masa de nitrógeno se debe calcular primero el reactivo limitante:

$n_{O_{2}} = \frac{m}{M} = \frac{25,6 g}{31,99 g/mol} = 0,800 moles$

$n_{CH_{3}NH_{2}} = \frac{4}{9}*0,800 moles = 0,356 moles$

De la ecuación anterior se tiene que la cantidad de moles de metilamina necesaria para reaccionar con 0,800 moles de oxígeno es 0,356 moles, y la cantidad de moles iniciales de metilamina es 0,5 moles, por lo tanto el reactivo limitante es el oxígeno.

Ahora, podemos calcular la masa de nitrógeno producida:

$n_{N_{2}} = \frac{2}{9}*n_{O_{2}} = \frac{2}{9}*0,8 moles = 0,18 moles$

$m_{N_{2}} = n_{N_{2}}*M = 0,18 moles*28,014 g/mol = 5,043 g$

Por lo tanto, se pueden producir 5,043 g de nitrógeno.

c) El redimiento de la reacción se puede calcular usando la siguiente fórmula:

$\% = \frac{R_{r}}{R_{T}}*100$

Donde:

$R_{r}$ : es el rendimiento real

$R_{T}$ : es el rendimiento teórico

$\% = \frac{3,5}{5,043}*100 = 69,4$

Entonces, el procentaje de rendimiento de la reacción es 69,4%.

Espero que te sea de utilidad!

5 0

2 years ago

is there a relationship between the relative age of stars in the number/identity of element is found within them

lukranit [14]

Yes. Stars use fusion to create nuclear energy, which is what makes them "alive". The older they are, the "bigger" the element in them is. Hydrogen turns into Helium, and when hydrogen is used up, the helium starts fusing into bigger elements. it stops at iron however. Once stars start fusing silicon to iron, it is doomed because it takes more energy than it gives off.

8 0

2 years ago

Which statement explains why large atoms are more reactive than small atoms?

Lunna [17]

A. Large atoms have valence electrons farther from the nucleus and lose them more readily, so they are more reactive than small atoms.

For example, the valence electron of a small atom like Li is tightly held. Lithium gently fizzes on the surface as it reacts with the water to produce hydrogen.

In contrast, the valence electron of a large atom like Cs is so loosely held that cesium exlodes on contact with water.

8 0

2 years ago