4p + 5O2 -> P4O10 ; the percent yield of PO4O10 when 6.20 g of phosphorus burns into excess oxygen is 67.0%. What is the yiel

Answer:

11.55 moles of CO2 gas result from 3.85 moles of propane.

15.4 moles of H2O result from 3.85 moles of propane.

b) 0.5176 moles of water result from 0.647 moles of oxygen gas.

0.1294 moles of propane are reacted.

Explanation:

a) Amount of propane = 3.85 moles

The reaction indicates that for every mole of propane, 3 moles of carbon dioxide are produced.

Thus, 3.85 moles of propane yield:

of carbon dioxide.

11.55 moles of CO2 gas result from 3.85 moles of propane.

The reaction also specifies that 1 mole of propane produces 4 moles of water.

Therefore, 3.85 moles of propane will generate:

of water.

15.4 moles of H2O result from 3.85 moles of propane.

b) Amount of oxygen gas = 0.647 moles

The reaction states that 5 moles of oxygen yield 4 moles of water.

As a result, 0.647 moles of oxygen will produce:

of water.

0.5176 moles of water result from 0.647 moles of oxygen gas.

The reaction indicates that 5 moles of oxygen gas interact with 1 mole of propane.

Then, 0.647 moles of oxygen will cause:

of propane to react.

0.1294 moles of propane are reacted.

Response:

The result is a pH of 7. I’m uncertain about the mechanics of Hydrochloric acid reactions

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

Answer:

The rate law for the decomposition reaction is:

The unit for the rate constant will be

Explanation:

The rate law can be expressed as:

..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

...[2]

[1] ÷ [2]

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

The unit for the rate constant will be:

The unit for the rate constant will be .

This procedure entails diluting the 12 molar HCl. To decrease the concentration, we must create an equation to determine how much of the 12M is needed for the 3.5M solution.

12 moles HCl 3.5 moles HCl
——————— = ———————
1 Liter of Soln ‘x’ Liters of Soln

Note that the ratio of 12 moles over 1 liter corresponds to 12 molar; thus, we maintain the original concentration of the HCl. By equating it to the 3.5 over ‘x’, we are still preserving the concentration.

After computation, we determine ‘x’ to be 0.292. This value indicates that within 0.292 liters of our 12 M HCl solution, there are 3.5 moles of HCl. Yet, we are not finished.

0.292 liters of 12 M HCl can create 1 liter of 3.5 M HCl, but the inquiry demands 1.5 liters. To achieve this, multiply 0.292 liters by 1.5, resulting in 0.4375, which denotes the quantity of 12 M HCl necessary to prepare a 1500 mL 3.5 M HCl solution.