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Karo-lina-s [1.5K]
2 years ago
14

Calculate the molar mass of NH4NO3. Molar mass of NH4NO3 = g/mol.

Chemistry
2 answers:
bija089 [108]2 years ago
4 0

The molar mass of NH4NO3 in g/mol is 80g/mol.

HOW TO CALCULATE MOLAR MASS:

The molar mass of a compound can be calculated by summing the atomic masses of its constituent elements.

In ammonium nitrate (NH4NO3), there are nitrogen, hydrogen, and oxygen elements.

  • Atomic mass of nitrogen = 14
  • Atomic mass of oxygen = 16
  • Atomic mass of hydrogen = 1

Molar mass of NH4NO3 = 14 + 1(4) + 14 + 16(3)

Molar mass of NH4NO3 = 80g/mol

  • Therefore, the molar mass of NH4NO3 in g/mol is 80g/mol.

Learn more about molar mass at: brainly.com/question/8101390?referrer=searchResults

krok68 [10]2 years ago
4 0

Answer:

80g/mol

Explanation:

edge 2022

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When the paramagnetic [co(cn)6] 4– ion is oxidized to [co(cn)6] 3– , the ion becomes diamagnetic. however, when the paramagnetic
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Answer:  
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not. 
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 Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it. 
 So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’. 
 You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not. 
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  Co3+:1s22s22p63s23p63d6 
 For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split. 
 More specifically, you can say that 
  a strong field ligand will produce a more significant splitting energy, Δ  a weak field ligand will produce a less significant splitting energy, Δ 
 Now, the spectrochemical series looks like this 
 http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm 
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6 0
3 years ago
opper(I) ions in aqueous solution react with NH 3 ( aq ) according to Cu + ( aq ) + 2 NH 3 ( aq ) ⟶ Cu ( NH 3 ) + 2 ( aq ) K f =
Y_Kistochka [10]

Answer:

55.373g/l

Explanation:

The dissolved amount of sparingly soluble salts is interlinked with a unitlesss quantity called as solubility product. It is a fixed quantity that only increases with the rise in temperatures and is used to predict the salting out of compounds. If the value of ionic product (Q) is larger than (Ksp), precipitation of compound occurs.

Given:

The solubility product of CuBr is 6.3×10−9.

The concentration of NH3 is 0.10 M.

Formula and Calculations:

The dissolution reaction (I) of CuBr is shown below.

The reaction showing dissolution of CuBr in NH3 is shown below.

The above reaction can be obtained by adding reaction (I) and (II) as shown below.

The equilibrium constants will get multiplied.

Suppose the solubility of CuBr is “s”.

It is given that concentration of NH3 is 0.10 M.

The equilibrium constant expression for the above reaction is as follows,

Here,

The concentration of pure solids is 1 M. Thus, the concentration of CuBr is 1 M.

As calculated, the value of Ksp is 396.9.

Substitute all the required values in above formula.  

On further solving above equation,

Therefore, the solubility of CuBr in ammonia is 0.386 M.

The formula to calculate solubility

Solubuility (g/l)= Molarity(M) x Molarmass

Chemistry homework question answer, step 2, image 10

The molar mass of CuBr is 143.45 g/mol.

The formula to calculate solubility in g/L is given below.

The molar mass of CuBr is 143.45 g/mol.

therefore,

solubility = 0.386M x 143.45g/mol

where (M = mol/l)

solubility = 55.375g/l

5 0
3 years ago
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