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musickatia [10]
3 years ago
7

how many moles of ammonium chloride are required for a complete reaction with 420 mol of calcium hydroxide

Chemistry
1 answer:
Oxana [17]3 years ago
3 0

3 KOH + H3PO3 = K3PO4 + 3 H2O  

moles KOH = 26.5 g/56.1 g/mol=0.472  

moles K3PO4 = 0.472/3 =0.157

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Explanation:

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What is the relationship between the degree of ionization and conductivity
LuckyWell [14K]
The higher the degree of ionization, the stronger the conductivity
5 0
3 years ago
2000 minutes to hours dimensional analysis
elena-14-01-66 [18.8K]

Answer:

33.33 h

Explanation:

You know that

1 h = 60 min

If you divide both sides by 60 min, you get the conversion factor: 1 h/60 min = 1.

If you divide both sides by 1 h, you get the conversion factor: 1 = 60 min/1 h.

Both are conversion factors because they both equal one and multiplying a measurement by one does not change its value.

You choose the conversion factor that gives you the correct dimensions for your answer. It must have the correct dimensions on top (in the numerator),

Thus, to convert 2000 min to hours, you use the conversion factor with “h” on the top.

\text{Time} = \text{2000 min} \times \dfrac{\text{1 h}}{\text{60 min}} = \textbf{33.33 h}

5 0
3 years ago
Which part(s) of the reaction 2 HD(g) ⇀↽ H2(g) + D2(g) will be favored by an increase in the total pressure (resulting in compre
Sergio [31]

Answer:

Neither side is favored

Explanation:

In a reaction at equilibrium in which reactants and products are gases, pressure is a factor that produces a shift of the equilibrium position. When the pressure is increased, the equilibrium is shifted to the side of the reaction with a fewer moles. In this reaction:

2 HD(g) ⇀↽ H₂(g) + D₂(g)

There is the same number of moles on both sides:

reactants side: 2 moles HD

products side: 1 mol H₂ + 1 mol D₂ = 2 moles

Therefore, the equilibrium position will not change.

7 0
3 years ago
The phenolic indicator (In-OH)has approximately the same pKa as a carboxylic acid. Which H is the most acidic proton in In-OH? C
Sidana [21]

The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.

The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).

The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).  

The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.

Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.    


4 0
3 years ago
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