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Butoxors [25]
4 years ago
13

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water? What mass of water is needed t

o prepare a 1.20 molal solution using 0.60 mol propyleneglycol? What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?
Chemistry
1 answer:
klemol [59]4 years ago
8 0

Answer:

1) The molality is 0.145 molal

2) We need 0.50 kg water

3) The molality is 0.245 molal

Explanation:

What is the molality of a solution in which 0.32 moles AlCl3 has been dissolved in 2,200 g water?

Molality = moles AlCl3 / mass water (in kg)

Molality = 0.32 moles / 2.2 kg water

Molality = 0.145 moles/kg = 0.145 molal

The molality is 0.145 molal

What mass of water is needed to prepare a 1.20 molal solution using 0.60 mol propyleneglycol?

Molality = moles propyleneglycol / mass water

1.20 = 0.60 / x kg

X = 0.60 / 1.2

X = 0.50 kg water

We need 0.50 kg water

What is the molality of a solution in which 0.145 mol CO2 is dissolved in 591 g water?

Molality = moles CO2 / mass water (in kg)

Molality = 0.145 moles / 0.591 kg

Molality = 0.245 molal

The molality is 0.245 molal

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What is the effect of substituting soluble metal salts such as NaCl and K2CO3 for HCl or (NH4)2CO3 in the qualitative analysis s
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8 0
3 years ago
What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) =&gt; 2
zavuch27 [327]

<u>Answer:</u> The mass of iron (II) oxide that must be used in the reaction is 30.37

<u>Explanation:</u>

The given chemical reaction follows:

6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, \frac{6}{635}\times 44.7=0.423mol of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37

7 0
4 years ago
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