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2 years ago

Use the model provided to select ALL of the examples of mixtures.

Chemistry

2 answers:

Serggg [28]2 years ago

6 0

Answer:

Explanation:

Ice tea

Salt water

Orange juice

ruslelena [56]2 years ago

3 0

Where is the model? I do know salt water is a mixture though

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A pure compound is found to be 40.0% carbon by mass, 6.73% hydrogen by mass, and 53.3% oxygen by mass. determine the empirical f

ratelena [41]

Since there is no weight, I would assume that this is a 100g of pure compound.
Okay so I would be changing the percentage to gram to solve for the mole.
So
40.0g C (1 mol C/12.01 g C) = 3.33 mol C
6.73g H (1 mol H/1.01 g H ) = 6.66 mol H
53.3g O (1 mol O/16.00 g O) = 3.33 mol O

With that, two of our moles is 3.33, so we consider that are our 1, as it is also the lowest. Therefore the empirical formula is CH2O

3 0

2 years ago

Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you

oee [108]

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

$\eta_{KClO_{3}} = \frac{m}{M}$

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

$\eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles$

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

$\eta_{O_{2}} = \frac{3}{2}*0.252 moles = 0.378 moles$

Finally, the mass of O₂ is:

$m = 0.378 moles*32 g/mol = 12.10 g$

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

6 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$