How many moles are in 8.35 g of H2O?

Zingerone, one of the flavor molecules in ginger,

MissTica

The molecular formula of the compound is C12H15O3 hence the molar mass of the compound is 207 g/mol.

We need to obtain the number of moles of carbon, hydrogen and oxygen in the compound;

Carbon = 24.91 g/44g/mol × 1 mole of carbon = 0.566 moles

Mass of carbon = 0.566 moles × 12 g/mol = 6.792 g

Number of moles of hydrogen = 6.522 g/18 g/mol × 2 moles = 0.725 moles

Mass of hydrogen = 0.725 moles × 1 g/mol = 0.725 g

Mass of oxygen = 10 - (6.792 g + 0.725 g) = 2.483 g

Number of moles of oxygen = 2.483 g/16 g/mol = 0.155 moles

Now we must divide through by the lowest number of moles;

C - 0.566/0.155 H - 0.725/0.155 O - 0.155/0.155

C - 4 H - 5 O - 1

The simplest formula is C4H5O Recall that the molar mass of the compound lies between 150.0 and 220.0 g/mol

4(12) + 5(1) + 16 = 69

Hence; n = 3 and the molecular formula of the compound is C12H15O3

The molar mass of the compound is; 12(12) + 15(1) + 3(16) = 207 g/mol

Learn more: brainly.com/question/15180604

3 0

3 years ago

On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.

Delicious77 [7]

Answer:

Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes.

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

3) Now locate and label the points:

(90, 0.9) ⇒ 90 mb, 0.9 ml

(100, 0.8) ⇒ 100 mb, 0.8 ml

(400, 0.2) ⇒ 400 mb, 0.2 ml

(600, 0.15) ⇒ 600 mb, 0.15 ml

(760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

5 0

3 years ago

Read 2 more answers

A mixture contains only nacl and al2(so4)3. a 1.45 g sample of the mixture is dissolved in water, and an excess of naoh is added

Aleks [24]

When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-

Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄

On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.

As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.

So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = $\frac{1.085 X 100}{1.45}$ =74.8 %.