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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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egoroff_w [7]

Answer:

<em>The</em><em> </em><em>Typical</em><em> </em><em>Oxidation</em><em> </em><em>number</em><em> </em><em>of</em><em> </em><em>Oxygen</em><em> </em><em>is</em><em> </em><em>–</em><em>2</em><em>.</em>

<em>And</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>question</em><em> </em><em>said</em><em>.</em><em>.</em><em>.</em><em> </em><em>its</em><em> </em><em>Oxidation</em><em> </em><em>changes</em><em> </em><em>to</em><em> </em><em>–</em><em>1</em><em> </em><em>when</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>in</em><em> </em><em>Peroxides</em><em> </em><em>and</em><em> </em><em>–</em><em>½</em><em> </em><em>w</em><em>h</em><em>e</em><em>n</em><em> </em><em>i</em><em>t</em><em>s</em><em> </em><em>i</em><em>n</em><em> </em><em>SuperOxides</em><em>.</em>

<em>Correct</em><em> </em><em>Answer</em><em> </em><em>:</em><em> </em><em>Option</em><em> </em><em>D</em><em>.</em>

7 0
2 years ago
A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it
patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0
3 years ago
If an atom can form more than one cation, the cation with the lowest charge ends with which set of letters?
Verdich [7]

Answer:

Answer is A 

Explanation:

Now we know that an atom wants to complete its outer shell while keeping electrons in pairs of two now in A there are four electrons which which can be ejected while in B will want to accept 3 electrons to complete its shell as ejecting five will take lot of energy similar case will be for C,D and E which would want to accept 2,1,0 electrons respectively

3 0
3 years ago
Read 2 more answers
If the outside of your flask is not dry when the first mass determination is made and it is dried for the second mass determinat
svlad2 [7]

Answer:

You'll experience a grater deviation

Explanation:

<em>You'll experience a greater deviation in your measurements, meaning your measures will have a bigger difference between them, and the greater these deviations the less accurate will be the measuring.</em> This happens mainly because you're not replicating the measurement with the exact same conditions, in one of them you'll have an extra mass from the water.

I hope you find this information useful and interesting! Good luck!

8 0
3 years ago
Titration Volume &amp; Concentration
polet [3.4K]

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

0,0171L*\frac{0,161moles}{L} = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}} = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

5,5062x10^{-3}molHCl*\frac{1L}{0,304mol} = 0,0181L≡18,1mL

I hope it helps!

4 0
3 years ago
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