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photoshop1234 [79]
3 years ago
10

Is the soup saturated with salt at 20°C? Explain why or why not.

Chemistry
1 answer:
yKpoI14uk [10]3 years ago
6 0
The salt doesn't dissolve at that temperature. heating aids in dissolving
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The electron configuration of calcium is
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20 Ca :

1s², 2s², 2p⁶, 3s², 3p⁶, 4s²<span>
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Answer D

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Please help!!! I’ll mark brainlist
Lerok [7]

Explanation:

they are closer and made of gas

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Which set of data would be best illustrated with a graph bar?
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Answer:

Graphs should be titled.

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3 years ago
What is the ratio of the areas of the signals in the h nmr spectrum of pentan-3-ol?
Damm [24]

The ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.

<h3>What is a NMR spectrum?</h3>

Nuclear magnetic resonance spectroscopy is a spectroscopy that shows the detailed structure and chemical environment of a chemical element.

Pentan-3-ol contain 12 hydrogen atoms. In H-NMR spectra, hydrogen atoms have same environment gives a signal.

There are 4 different kinds of signals due of the 4 different environment experienced by these 12 hydrogens.

Thus, the  ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.

Learn more about NMR spectrum

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8 0
2 years ago
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
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