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Rasek [7]
3 years ago
8

What is the molality of a solution prepared by dissolving 150.0 g C6H12O6 in 600.0 g of H2O?

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

50/36 = 25/18

Explanation:

Solution at attachment box

Molality = mole of dissolvable (this question glucose) / kg of water

Masja [62]3 years ago
3 0

Answer:

Molality = 1.38 mol/Kg

Explanation:

Given data:

Molality of solution = ?

Mass of C₆H₁₂O₆ = 150.0 g

Mass of water = 600.0 g (600 g ×1 kg/1000 g= 0.6 Kg)

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 150.0 g/180.16 g/mol

Number of moles = 0.83 mol

Molality:

Molality = moles of solute / kg of solvent

Molality =  0.83 mol /0.6 Kg

Molality = 1.38 mol/Kg

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3 years ago
Metals react with oxygen to give oxides with the general formula MxOy. What is a balanced chemical equation for the reaction of
katovenus [111]

Answer: 4Fe+3O_2\rightarrow 2Fe_2O_3

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

Here iron is having an oxidation state of +3 called as Fe^{3+} cation and oxide O^{2-} is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral Fe_2O_3.

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

2Fe+3O_2\rightarrow 2Fe_2O_3

8 0
3 years ago
Read 2 more answers
What is that the theoretical yield of aluminum oxide I if 3.20 mol of aluminum metal is exposed to 2.70 mole of oxygen
photoshop1234 [79]

Answer:

163.2g

Explanation:

First let us generate a balanced equation for the reaction. This is shown below:

4Al + 3O2 —> 2Al2O3

From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.

From the equation,

4moles of Al produced 2moles of Al2O3.

Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.

Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:

Mole of Al2O3 = 1.6mole

Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol

Mass of Al2O3 =?

Number of mole = Mass /Molar Mass

Mass = number of mole x molar Mass

Mass of Al2O3 = 1.6 x 102 = 163.2g

Therefore the theoretical of Al2O3 is 163.2g

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