According to kinetci moleculartheory, which of the following would not be onsidered an ideal gas

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Answer 1

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Answer:

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Ammonia, NH3, is used as a refrigerant. At its boiling point of –33 oC, the standard enthalpy of vaporization of ammonia is 23.3 kJ/mol. How much heat is released when 50.0 g of ammonia is condensed at –33 oC?–0.466 kJ–7.94 kJ–36.6 kJ–68.4 kJ–1.17 x 103 kJ

Answers

Answer:

-68.4 kJ

Explanation:

The standard enthalpy of vaporization = 23.3 kJ/mol

which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).

To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.

This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.

Thus, Q = -23.3 kJ/mol

Where negative sign signifies release of heat

Given: mass of 50.0 g

Molar mass of ammonia = 17.034 g/mol

Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles

Also,

1 mole of ammonia when condenses at -33 °C releases 23.3 kJ

2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ

Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.

Final answer:

The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.

Explanation:

To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.

Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:

2.94 moles x 23.3 kJ/mol = 68.4 kJ

Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.

Learn more about Enthalpy of Vaporization here:

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Of the following, which element has the highest first ionization energy?Li
Cs
At
F

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There are free tutor on my website just got to freetutor.com

I think it’s Li but I don’t totally know

Identify the sample and analyte in each of the scenarios.(a) Determination of the lead content in paint. lead paint
(b) Analysis of the nitrate content of soil near a local water source. soil nitrate
(c) Measurement of the citric acid found in a lime.

Identify the following as either sample or analyte.
(1) lead
(2) paint chips
(3) soil
(4) nitrate
(5) lime wedge
(6) citric acid

Answers

Answer:

a) Analyte: lead. Sample: paint.

b) Analyte: nitrate. Sample: soil.

c) Analyte: citric acid. Sample: Lime

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Explanation:

A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.

Thus:

a) Analyte is lead while you must take a sample of paint to analyze this lead.

b) Analyte is the nitrate while sample must be soil.

c) Analyte is citric acid and lime is the sample

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

270kJ are needed to keep a 75 watt light bulb burning for 1 hour. How many kilocalories are needed to burn this light bulb for 3 hours?

Answers

Answer:

For 1 hour 75w light bulb requires 270 kj for burning

for 3 hours 75 w light bulb requires 270*3 = 810kj for burning

Explanation:

How many milliliters of a 70.0 ml solution of 1.51 m bacl2 must be used to make 12.0 ml of a solution that has a concentration of 0.300 m cl–?

Answers

$1.43 \; \text{ml}$

Explanation

Number of moles of chloride ions in the final solution: $n = c \cdot V = 0.012 * 0.300 = 0.0036 \; \text{mol}$

Each mole of barium chloride contains two moles of chloride ions. Thus the concentration of chloride ions in the initial 1.51 M barium chloride solution: $\begin{array}{lll} c(\text{Cl}^(-)) &=& 2 * c(\text{BaCl}_2)\n& =& 2.52 \; \text{mol} \cdot \text{L}^(-1) \n & = &0.00252 \; \text{mol}\cdot \text{mL}^(-1) \end{array}$

Therefore

$\begin{array}{lll} V & = & n /c \n & = & 0.0036 / 0.00252\n & = & 1.43 \; \text{mL}\end{array}$

To test Döbereiner’s idea, predict:(a) The boiling point of HBr from the boiling points of HCl (- 84.9°C) and HI (-35.4°C) (actual value = -67.0°C)
(b) The boiling point of AsH₃ from the boiling points of PH₃ (- 87.4°C) and SbH₃ (-17.1°C) (actual value = -55°C)

Answers

Answer:

a) Approximate boiling point of HBr = -60.15 °C

b) Approximate boiling point of AsH₃ = -52.25 °C

Explanation:

Döbereiner stated that some elements could be arranged in groups of 3 similar elements ( known as "triads) , and the element of the middle ( elements are ordered with respect to their atomic mass) would have properties between the other 2 ( the average value)

a) In the first case the triad would be the halogen triad ( Cl , Br and I ) . And according to Döbereiner , the boiling point of HBr should be the average of HCl and HI . Therefore

Approximate boiling point of HBr = [(- 84.9°C) + (-35.4°C)]/2 = -60.15 °C

b) Simmilarly for AsH₃ , PH₃ and SbH₃ , the boiling point of AsH₃ would be

Approximate boiling point of AsH₃ = [(- 87.4°C) + (-17.1°C)]/2 = -52.25 °C

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