Help with the chart please

If a mixture looks smooth and the same throughout it is probably what

gulaghasi [49]

Answer:
If a mixture looks smooth and the same throughout it is probably Homogeneous.

Explanation:
Mixture is the combination of different compounds which are unreactive to each other.

Mixture are classified as ...

Solutions; in which the mixed compounds are thoroughly mixed and cannot be distinguished from each other and are said to be homogeneous. In solutions the size of solute is very small (i.e. Less than 1 nm).

Colloids; in which the solute is homogeneous visually but heterogeneous microscopically. The size of particles in this case is between 1 nm to 1 μm.

Suspensions; in which the mixture is heterogeneous, the particle size is greater than 1 μm and settles down (precipitation) under the influence of gravity.

4 0

3 years ago

One glass of water is 85 degrees F and another is 40 degrees F. When an Alka Seltzer tablet is dropped into each glass, at the s

Andreyy89

Answer:

the cold will get hot and the hot will get cold

Explanation:

7 0

3 years ago

Tema: Métodos de Separação de Misturas – Homogêneas e Heterogêneas;

seraphim [82]

Answer:

fjskeowkcnekvo Dee five votes come vote for dog even r

4 0

2 years ago

What would be the new volume if the temperature on 800 mL of gas is changed

dolphi86 [110]

I think you divide something

8 0

2 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago

Help with the chart please​

Help with the chart please