Quick question: Are water molecules attracted to each other because of the polar nature of water?

For each of the following solutions, calculate [OH- ] from [H3O + ] or [H3O + ] from [OH- ]. Classify each solution as acidic, b

Yuri [45]

The case is acidic since calculated pH is below 7 that is 2.3

The hydroxide and hydrogen ion levels in a solution are the sources for calculating pOH and pH levels, respectively, and are the basis of whether a solution is acidic or basic

[H₃O+]=5.3×10⁻³M

To determine the hydroxide ion concentration, we use:

[OH−][H3O+]=1×10⁻¹⁴

We isolate the hydroxide ion concentration:

[OH−]=1×10−145.3×10⁻³M

Solving this, we obtain:

[OH−]=1.9×10−¹²M

We can determine whether the solution is acidic or basic by calculating the pH level:

pH=−log[H3O+]

We substitute in the hydronium ion concentration given:

pH=−Log(5.3×10−3)pH=2.3

Since the pH is below 7, the solution is acidic

Learn more about hydroxide ion here-

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4 0

2 years ago

The chemical equation below shows the combustion of propane (C3H8). C3H8 + 5O2 --> 3CO2 + 4H2O What mass of O2, in grams, is

Over [174]

Answer:

plk

Explanation:

7 0

3 years ago

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Tamiku [17]

Answer:

The explosive decomposition of 98.4 kg of ammonium nitrate produces 58498.8 L of nitrogen, 29249.4 L of oxygen and 116997.6 L of water vapor.

Explanation:

To find how many liters of gas are formed from the explosive decomposition of ammonium nitrate it is necessary to follow these steps (check the attachment for better understanding):

1st) Balance the equation:

Write the decomposition equation and then find the correct coefficients to make sure that it goes according to the "Law of conservation of mass" (the mass of the reactants side must be equal to the mass of the products side). So, 2 moles of ammonium nitrate produces 2 moles of nitrogen, 1 mole of oxygen and 4 moles of water vapor.

2nd) Find the Ammonium nitrate molar mass:

The ammonium nitrate mass it is calculated by adding de molar mass of each atom that forms the ammonium nitrate molecule. You can find the elements molar mass in the Periodic Table.

In this example I decided to round the number to simplify tha calculus, for example: the oxygen molar mass in the periodic table is 15.9994 but I use 16. You can use the complete number if you want.

By doing this, the ammonium nitrate molar mass is 80 g/mol.

The statement says that there is 98.4 kg of ammonium nitrate. In ordder to use the same units in all the calculus sometimes it is usefull to convert the kg to g, so it is the same as 98400g. You can do it the other way around if you prefer (g to kg).

3rd) Find the number of moles of each gases and aqua vapor formed:

It is important to know the amount of each compound formed by the decomposition reaction, that's why we need to pay attention to the coefficients of the balanced reaction.

The amount of each compound is easily found by using the "rule of three".

To use the rule of three we need to think using the balanced reaction so:

If 160g (2 moles) of ammonium nitrate produces 2 moles of nitrogen gas, the 98400g that we have of ammonium nitrate will produce an X amount of nitrogen gas. With this information we multiply 98400g by 2 moles and then we divide the result by 160g. The final result it is 1230 moles of nitrogen.

In the same way we use the rule of three to calculate the number of moles of oxygen and water.

4th) Find the liters (volume) of each gas and aqua vapor formed:

Finally, to find the liters from the number of moles, it is necessary to apply the "Ideal gases law", that relates the pressure (atm), volume (L), moles number and temperature (Kelvin) with the R gas constant in the formula:

PxV = nxRxT

It is important to use the correct units because the R gas constant is equal to 0.082 atm.L/mol.K.

As we need to calculate the liters (volume) we pass the pressure dividing to the other side and then we just have to replace the information:

V = (nxRxT)/P

As you can see in the attachment, doing this last step for each compound, we can find the liters produced of them.

8 0

3 years ago

Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite un

maw [93]

Answer:

7.23 J

Explanation:

Step 1: Given data

Mass of graphite (m): 566.0 mg

Initial temperature: 5.2 °C

Final temperature: 23.2 °C

Specific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹

Step 2: Calculate the energy required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)

Q = 7.23 J

6 0

3 years ago

What two pieces of evidence did van Helmont use to reach his conclusion

Sav [38]

Explanation:

He weighed silver, dissolved it in acid, and then recovered all the original silver by reacting the solution with copper. He also showed, by using iron to recover the copper, that this displacement of one metal from its salt by using a second metal was not because of transmutation, as many had held.

6 0

2 years ago