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2 years ago

Help! Locks at 12

Chemistry

2 answers:

marta [7]2 years ago

4 0

Answer is #4 (neutrons)

Explanation:

Total mass before the reaction is 236 amu. The total mass of the two elements is 233 amu. This means that there are 3 amus carried by element X. Looking at the total atomic number of the two elements after reaction, it's 92, the same as the one at the beginning. This means that element X has zero atomic number, which could only be a neutron.

In-s [12.5K]2 years ago

3 0

Answer: is 4

Explanation:

I was doing the same one

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1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data

earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v = 1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15 K

v₂ = 0.773 T₂ = 273.15 K

v₃ = 1.057 T₃ = 373.15 K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533 = 0.00284×T - 0.001167

v = 0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 = 0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411 -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

3 0

3 years ago

Describe radiation in a full sentence

Mandarinka [93]

Answer:

Radiation is energy that comes from a source and travels through space at the speed of light.

3 0

3 years ago

An element with the smallest anionic (negative-ionic) radius would be found on the periodic table in

Advocard [28]

The correct answer for, An element with the smallest anionic (negative-ionic) radius would be found on the periodic table in, is <span>Group 17, Period 2.</span>

5 0

3 years ago

A wave of infrared light has a speed of 6 m/s and a wavelength of 12 m. What is the frequency of this wave?

andreev551 [17]

Answer: The frequency of the wave is 0.5 hertz.

Explanation:

$\nu =\frac{c}{\lambda }$

$\nu$ = Frequency of the wave

$c$ = speed of the light in m/s

$\lambda$ = Wavelength of the wave.

Here in question we are given with speed of the infrared light. So, we will replace the value of speed of light(c) from the given value of the speed of the infrared light.

Speed of infrared light = 6 m/s

$\nu =\frac{\text{speed of infrared light}}{\lambda }=\frac{6 m/s}{12 m}=0.5 hertz$

The frequency of the wave is 0.5 hertz.

7 0

3 years ago

Read 2 more answers

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m