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KIM [24]
4 years ago
15

Which of these elements is unlikely to have a reaction with any element or compound?

Chemistry
1 answer:
Kay [80]4 years ago
8 0
Argon is a noble gas. Argon has a full outer shell. This makes it so that it does not need to react with any of the other elements to be stable.


With Rubidium and Cobalt its a whole different story.

I hope that helps!
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Según la cinética química para que una reacción ocurra, los átomos o moléculas deben
kkurt [141]

Answer:

solo I

Explanation:

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4 0
3 years ago
Given that RH= 2.18 x 10⁻¹⁸J, 1 nm = 1 x 10⁻⁹m, h = 6.63 x 10⁻³⁴J·s, and c = 3.00 x 10⁸m/s:
Bas_tet [7]

Answer:

The wavelength the light emitted by a hydrogen atom during a transition is 1006 nm.

Explanation:

By using Rydberg's Equation we cab determine the wavelength of the light:

\Delta E=R_H\times Z^2\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\Delta E = Energy difference

R_H = Rydberg's Constant

n_f = Final energy level

n_i= Initial energy level

We have : n_i=7,n_f=3 , Z = 1

R_H=2.18\times 10^{-18} J

\Delta E=2.18\times 10^{-18} J\times 1^2\left(\frac{1}{7^2}-\frac{1}{3^2} \right )

\Delta E=1.9773\times 10^{-19} J

Now by using Plank's equation we can determine the wavelength of the light emitted.

E=\frac{hc}{\lambda }

E = Energy of the emitted light

h = Planck's constant = 6.63\times 10^{-34} Js

c = speed of light = 3.00\times 0^8 m/s

For the given transition the energy of the light = E

E =1.9773\times 10^{-19} J

\lambda=\frac{hc}{E}=\frac{6.63\times 10^{-34} Js\times 3.00\times 0^8 m/s}{1.9773\times 10^{-19} J}

\lambda =1.006\times 10^{-6} m =1.006\times 10^{-6}\times 10^9=1006 nm

The wavelength the light emitted by a hydrogen atom during a transition is 1006 nm.

3 0
3 years ago
A 6.22-kg piece of copper metal is heated from 20.5 °C to 324.3 °C. The specific heat of Cu is 0.385 Jg-1°C-1.a.Calculate the he
iren2701 [21]

Answer:

a) 727.5 kJ

Explanation:

Step 1: Data given

Mass of the piece of copper = 6.22 kg

Initial temperature of the copper = 20.5 °C

Final temperature of the copper = 324.3 °C

Specific heat of copper = 0.385 J/g°C

Step 2:

Q = m*c*ΔT

⇒ with Q = heat transfer (in J)

⇒ with m = the mass of the object (in grams) = 6220 grams

⇒ with c = the specific heat capacity = 0.385 J/g°C

⇒ with ΔT = T2 -T1 = 324.3 -  20.5 = 303.8

Q = 6220 grams * 0.385 J/g°C * 303.8 °C

Q = 727509.9 J = 727.5 kJ

b) This heat capacity is the heat capacity given for a copper at a temperature of 25°C

8 0
3 years ago
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