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Norma-Jean [14]
3 years ago
6

16. Which of the following are examples of ionic compounds? Select all that apply.

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

KBr, FeF3

Explanation:

Potassium ion (K+) + bromide ion (Br-) make the ionic compound KBr

Iron (III) ion (Fe3+) + 3 fluoride ions (F-) make the ionic compound FeF3

NH3 has covalent bonds so it isn't ionic

Mg12 is an elemental compound

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Answer:

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Explanation:

They always fought over petty things, but at the end, they did end up together!

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3 years ago
Helium is most likely to behave as an ideal gas when it is under(1) high pressure and high temperature(2) high pressure and low
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Helium is most likely to behave as an ideal gas when it is under 2) High pressure and low temperature. 
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4 years ago
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which of the following makes ice cores so useful in determining the composition of the atmosphere in the past?
iVinArrow [24]
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3 years ago
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A decorative "ice" sculpture is carved from dry ice (solid CO2) and held at its sublimation point of –78.5°C. Consider the proce
Leno4ka [110]

Answer:

The answers to the questions are;

a. The entropy of sublimation for carbon dioxide (the system) is  

134.07 J/Kmol.

b. The entropy of the universe for this reversible process is 376 J/K.

Explanation:

Entropy of sublimation is the entropy change experienced following the transformation of a mole of solid to vapor at  the temperature where the sublimation is taking place

a. We note that the mass of the solid CO₂ = 389 g

Molar mass of CO₂ = 44.01 g/mol

Number of moles of CO₂ in the sculpture = Mass/(Molar mass)

= (389 g)/(44.01 g/mol) = 8.84 Moles

Entropy of sublimation is given by

ΔS_{sublimation} = S_{vapor} - S_{solid} = \frac{\Delta H_{sublimation}}{T}

Where:

ΔH_{sublimation}  = 26.1 KJ/mol

T = Temperature = –78.5°C = ‪194.65‬ K

Therefore the amount of heat required to cause the 389 g of dry ice to sublime =    26.1 KJ/mol  × 8.84 Moles = 230.695 KJ

Therefore the entropy of sublimation = ΔS_{sublimation} = \frac{230.695 KJ}{194.65 K}

= 1.185 KJ/K

= 1185 J/K = 1185/8.84 J/Kmol = 134.07 J/Kmol

b. The entropy of the universe is given by;

ΔS_{universe} = \Delta S_{system} + ΔS_{surrounding}  

If the heat absorbed by the system is the same as the heat given off by the surrounding, then we have;

ΔS_{universe} = \frac{Q}{ T_{system}}  -\frac{Q}{T_{surrounding}}  

                =1.185 KJ/K - -\frac{230.695 KJ}{285.15K} = 1.185 KJ/K - 0.809 KJ/K = 0.376 KJ/K

= 376 J/K.

7 0
3 years ago
Question 6
kompoz [17]

Considering the combined law equation, the new temperature is -244.56 °C or 28.44 K.

<h3>Boyle's law</h3>

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant: if the pressure increases, the volume decreases while if the pressure decreases, the volume increases.

Mathematically, this law states that the multiplication of pressure by volume is constant:

P×V=k

<h3>Charles's law</h3>

Charles's law states that the volume is directly proportional to the temperature of the gas: if the temperature increases, the volume of the gas increases while if the temperature of the gas decreases, the volume decreases.

Mathematically, Charles' law states that the ratio of volume to temperature is constant:

\frac{V}{T} =k

<h3>Gay-Lussac's law </h3>

Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature: increasing the temperature will increase the pressure, while decreasing the temperature will decrease the pressure.

Mathematically, Gay-Lussac's law states that the ratio of pressure to temperature is constant:

\frac{P}{T} =k

<h3>Combined law equation</h3>

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

\frac{PxV}{T} =k

Considering an initial state 1 and a final state 2, it is fulfilled:

\frac{P1xV1}{T1} =\frac{P2xV2}{T2}

<h3>New temperature</h3>

In this case, you know:

  • P1= 1 atm= 760 mmHg
  • V1= 15 L
  • T1= -30 °C= 243 K (being 0 °C= 273 K)
  • P2= 58 mmHg
  • V2= 23 L
  • T2= ?

Replacing in the combined law equation:

\frac{760 mmHgx15 L}{243 K} =\frac{58 mmHgx23 L}{T2}

Solving:

T2x\frac{760 mmHgx15 L}{243 K} =58 mmHgx23 L

T2 =\frac{58 mmHgx23 L}{\frac{760 mmHgx15 L}{243 K}}

<u><em>T2= 28.44 K= -244.56 °C</em></u>

Finally, the new temperature is -244.56 °C or 28.44 K.

Learn more about combined law equation:

brainly.com/question/4147359

#SPJ1

6 0
2 years ago
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