A ___________________________ is made of two or more different substances.

Comment on whether each of the following is a homogeneous mixture or a heterogeneous mixture: (a) air in a closed bottle, (b) ai

Colt1911 [192]

Explanation:

A mixture in which there is uniform distribution of solute particles into the solvent is known as a homogeneous mixture.

For example, sugar dissolved in water is a homogeneous mixture.

On the other hand, a mixture in which there is uneven distribution of solute particles into the solvent is known as a heterogeneous mixture.

For example, sand present in water is a heterogeneous mixture.

Comment on given situations will be as follows.

(a) Air in a closed bottle - It is a homogeneous mixture because there will be even distribution of other gases that are present in air.

(b) Air over New York City - It is a heterogeneous mixture because there will be presence of some dust particles, fog or smoke into the air. Distribution of all these particles will be uneven. This will make air over New York City heterogeneous in nature.

8 0

3 years ago

3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 m

IrinaK [193]

Answer:

$49.45~%$

Explanation:

In this case, we have to start with the chemical reaction:

$C_6H_1_2O~->~C_6H_1_0~+~H_2O$

So, if we start with 10 mol of cyclohexanol ( $C_6H_1_2O$ ) we will obtain 10 mol of cyclohexanol ( $C_6H_1_0$ ). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:

$(6*12)+(10*1)=82~g/mol$

With this value we can calculate the grams:

$10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0$

Now, we have as a product 500 mL of $C_6H_1_0$ . If we use the density value (0.811 g/mL). We can calculate the grams of product:

$500~mL\frac{0.811~g}{1~mL}=405.5~g$

Finally, with these values we can calculate the yield:

$%~=~\frac{405.5}{820}x100~=~49.45%$ %= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

6 0

3 years ago

How many atoms of phosphorus are in 7.00 mol of copper(II) phosphate?

Rama09 [41]

Answer: The number of phosphorus atoms in given amount of copper(II) phosphate is $8.431\times 10^{24}$

Explanation:

We are given:

Moles of copper(II) phosphate $Cu_3(PO_4)_2$ = 7.00 mol

1 mole of copper(II) phosphate contains 3 moles of copper, 2 moles of phosphorus and 8 moles of oxygen atoms

Moles of phosphorus in copper(II) phosphate = $(2\times 7.00mol$

According to the mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of particles

So, 7.00 moles of copper(II) phosphate will contain = $(2\times 7\times 6.022\times 10^{23}=8.431\times 10^{24}$ number of phosphorus atoms.

Hence, the number of phosphorus atoms in given amount of copper(II) phosphate is $8.431\times 10^{24}$

3 0

3 years ago

(a)² + (b)² = ? help please

spayn [35]

answer:

a^2+b^2= (a+b)² -2ab

3 0

3 years ago

Redo the experiment by clicking on Reset Experiment. Add Ca(NO3)2 to 40 g of Na2CO3 and determine at what point the masses of th

Radda [10]

Answer : The mass of $Ca(NO_3)_2$ added must be, 61.9 grams

Explanation : Given,

Mass of $Na_2CO_3$ = 40 g

Molar mass of $Na_2CO_3$ = 105.9 g/mol

First we have to calculate the moles of $Na_2CO_3$ .

$\text{ Moles of }Na_2CO_3=\frac{\text{ Mass of }Na_2CO_3}{\text{ Molar mass of }Na_2CO_3}=\frac{40g}{105.9g/mole}=0.378moles$

Now we have to calculate the moles of $MgO$

The balance chemical reaction will be:

$Ca(NO_3)_2+Na_2CO_3\rightarrow CaCO_3+2NaNO_3$

From the balanced reaction we conclude that

As, 1 mole of $Na_2CO_3$ react with 1 mole of $Ca(NO_3)_2$

So, 0.378 mole of $Na_2CO_3$ react with 0.378 mole of $Ca(NO_3)_2$

Now we have to calculate the mass of $Ca(NO_3)_2$

$\text{ Mass of }Ca(NO_3)_2=\text{ Moles of }Ca(NO_3)_2\times \text{ Molar mass of }Ca(NO_3)_2$

Molar mass of $Ca(NO_3)_2$ = 164 g/mol

$\text{ Mass of }Ca(NO_3)_2=(0.378moles)\times (164g/mole)=61.9g$

Thus, the mass of $Ca(NO_3)_2$ added must be, 61.9 grams

6 0

4 years ago