If 7.50 g of the unknown compound contained 0.250 mol of c and 0.500 mol of h, how many moles of oxygen, o, were in the sample?

Answer:

• Please refer to the attached image for the graph with labeled axes and points.

Explanation:

This is a guide to fulfilling the instructions along with essential notes for understanding how to create such graphs:

1) The horizontal axis should indicate pressure ranging from 0 mb to 760 mb, while the vertical axis corresponds to volume ranging from 0 to 1 mL.

The x-axis captures the independent variable, and the y-axis records the dependent variable. Both axes must be accurately labeled, showing the variable names and their respective units.

In this context, the origin, (0,0), signifies the intersection of the axes at a pressure of 0 mb and a volume of 0.0 milliliters.

2) Allocate values for the divisions on the axes to maximize the usage of space on both.

An effective graph aims to utilize the entire space on both axes; for this, identify the maximum values for pressure and volume, and determine the corresponding marks.

The pressure range along the x-axis is [90, 760 mb], suggesting large divisions of 100 mb, with the farthest right mark at 800 mb. You can then subdivide each 100 mb interval into 10 smaller sections, using small divisions of 10 mb (my example employs 4 sections of 25 mb, but 10 mb is preferable).

The volume's range for the vertical axis is [0.1, 0.8], so it’s best to use divisions set at 0.1 ml.

3) Next, identify and label the points as follows:

• (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 represented as (x, y) are referred to as ordered pairs, indicating that the sequence is significant: the first number denotes the independent variable whereas the second denotes the dependent variable.

Thus, for the point (90, 0.9), 90 indicates a pressure of 90 mb and 0.9 indicates a volume of 0.9 ml.

To find (600, 0.15), since the horizontal increments are valued at 0.1, you should place the second coordinate of the point between the marks corresponding to 0.1 and 0.2 ml.

This allows you to accurately plot each point on the graph.

The aqueous solution of Na3PO4 is described as "strongly basic."

<span>4.3065 g To begin with, consult the atomic masses for each involved element. Atomic weight of Calcium = 40.078 Atomic weight of Carbon = 12.0107 Atomic weight of Hydrogen = 1.00794 Atomic weight of Oxygen = 15.999 Atomic weight of Sulfur = 32.065 Next, compute the molar masses of both reactants and the product. Molar mass H2SO4 = 2 * 1.00794 + 32.065 + 4 * 15.999 = 98.07688 g/mol Molar mass CaCO3 = 40.078 + 12.0107 + 3 * 15.999 = 100.0857 g/mol Molar mass CaSO4 = 40.078 + 32.065 + 4 * 15.999 = 136.139 g/mol The balanced equation for the reaction between H2SO4 and CaCO3 is: CaCO3 + H2SO4 ==> CaSO4 + H2O + CO2 Thus, 1 mole each of CaCO3 and H2SO4 is necessary to generate 1 mole of CaSO4. Let's check the amount of moles we have for CaCO3 and H2SO4. CaCO3: 3.1660 g / 100.0857 g/mol = 0.031632891 mol H2SO4: 3.2900 g / 98.07688 g/mol = 0.033545113 mol H2SO4 is in slight excess, therefore CaCO3 is the limiting reactant, suggesting we can expect 0.031632891 moles of product. To find the mass, multiply the number of moles by the molar mass calculated previously. 0.031632891 mol * 136.139 g/mol = 4.306470148 g Given that we have 5 significant figures from our data, we round the final result to 5 figures, yielding 4.3065 g</span>

The response is:

No, the equation is not balanced. Neither the Nitrogen (N) nor the Hydrogen (H) are in balance!

Here's the reasoning:

⓵ A properly balanced chemical equation means that the quantity of atoms on the reactants side matches that on the products side.

→ The equation lacks balance because there are 2 Nitrogen atoms and 2 Hydrogen atoms on the reactants side. In contrast, on the products side, there is only 1 Nitrogen atom and 4 Hydrogen atoms. Thus, the number of atoms on each side is not consistent!

Hopefully, this clarification is helpful; feel free to reach out if you have any further questions! ☻