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Serga [27]
3 years ago
12

At standard pressure, ammonia melts at 195 K and boils at 240 K. If a sample of ammonia at standard pressure is cooled from 200

K down to absolute zero, what physical constants are needed to calculate the change in enthalpy? I) the heat capacity of ammonia(s) II) the heat capacity of ammonia(ℓ) III) the heat capacity of ammonia(g) IV) the enthalpy of fusion of ammonia V) the enthalpy of vaporization of ammoni
Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Answer:

I) the heat capacity of ammonia(s)

II) the heat capacity of ammonia(ℓ)

IV) the enthalpy of fusion of ammonia

Explanation:

Initially, ammonia at 200 K is liquid. To calculate the change of enthalpy from 200 K to 195 K (melting point) we need to know the heat capacity of ammonia(ℓ).

At 195, ammonia is in the transition from liquid to solid (solidification). To calculate the change of enthalpy in that process we need to know the enthalpy of solidification of ammonia, which has the same value but opposite sign to the enthalpy of fusion of ammonia.

From 195 K to 0 K, ammonia is solid. To calculate the change of enthalpy in that process we need to know the heat capacity of ammonia(s).

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PLEASE SOMEONE HELP ME WITH THIS!
Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877  and 4) 1.04x10^{3\\}

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x  \frac{1 L}{1000 mL} = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x \frac{10 dg}{ 1 g} = 1.291 dg

For the next exercises, you need to follow some rules:

1.  All numbers  that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062  - 638.1848  = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.  

1.04x10^{3}

4 0
3 years ago
What is the Kb expression for aniline (C6H5NH2)?
Dimas [21]
This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]

hope this helps:)

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4 0
3 years ago
Read 2 more answers
Which of the following pairs of elements is most likely to share one or more pairs of electrons between their nuclei?
Nana76 [90]
Sulfur and chlorine. Explanation: A covalent bond is formed by two non-metals with similar electronegativities. As a consequence, they share one or more pairs of electrons between their nuclei
3 0
2 years ago
What is the binding energy of a nucleus that has a mass defect of 5.81*10-^29 kg
IrinaVladis [17]

Answer:

Choice A: Approximately 5.23 \times 10^{-29} joules.

Explanation:

Apply the famous mass-energy equivalence equation to find the energy that correspond to the \rm 5.81\times 10^{-29} kilograms of mass.

E = m \cdot c^{2},

where

  • E stands for energy,
  • m stands for mass, and
  • c is the speed of light in vacuum.

The speed of light in vacuum is a constant. However, finding the right units for this value can simplify the calculations a lot. What should be the unit of c?

The mass given is in the appropriate SI unit:

Mass is in kilograms.

Thus, proceed with the speed of light in SI units. The SI unit for speed is meters per second. For the speed of light, c \approx \rm 3.00\times 10^{8}\;m\cdot s^{-1}.

Apply the mass-energy equivalence:

\begin{aligned} E &= m \cdot c^{2} \\ &= \rm 5.81\times 10^{-29}\; kg \times {\left(3.00\times 10^{8}\; m\cdot s^{-1}\right)}^{2}\\ &\approx \rm 5.23\times 10^{-12}\;kg\cdot m^{2}\cdot s^{-2} \end{aligned}.

The unit of energy is not in joules. Don't be alerted. Consider the definition of a joule of energy. One joule is the work done on an object when a force of one newton acts on the object in the direction of the force through the distance of one meter. (English Wikipedia.)

\rm 1\; J = 1\; N \times 1\; m.

However, a force of one newton is defined as the force required to accelerated an object with a mass of one kilogram (not gram) at a rate of one meter per second squared. (English Wikipedia.)

\begin{aligned}\rm 1\; J &= \rm 1\; N \times 1\; m\\ & = \rm \left(1\; kg\times 1\; m\cdot s^{-2}\right)\times 1\; m\\ &= \rm 1\; kg \cdot m^{2}\cdot s^{-2}\end{aligned}.

In other words, the mass defect here is also \rm 5.23\times 10^{-12}\; J.

7 0
3 years ago
Read 2 more answers
Why doesnt ammonia support combustion?
earnstyle [38]
We need the reading for this I think
4 0
2 years ago
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