If the number is not a proper coefficient, how would you make it one?

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

How does the concentration of the substrate in an enzyme-controlled chemical reaction change over time?

Zinaida [17]

Answer:

The rate of a chemical reaction is directly proportional to the concentration. The rate of a chemical reaction increases as the substrate concentration increases and thus, concentration of the substrate in an enzyme-controlled chemical reaction increases with time.

Explanation:

The rate of a chemical reaction is directly proportional to the concentration.

The reaction rate increases with increasing substrate concentration, but levels off at a much lower rate. By increasing the enzyme concentration, the maximum reaction rate greatly increases.

Generally, the rate of a chemical reaction increases as the substrate concentration increases, and thus, concentration of the substrate in an enzyme-controlled chemical reaction increases with time.

5 0

3 years ago

Read 2 more answers

A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 1.80 fg (where a femtogram

egoroff_w [7]

Answer:

Δx ≥ 1.22 *10^-10m

Explanation:

Step 1: Data given

The E. coli bacterial cell has a mass of 1.80 fg ( = 1.80 * 10^-15 grams = 1.80 * 10^-18 kg)

Velocity of v = 8.00 μm/s (= 8.00 * 10^-6 m/s)

Uncertainty in the velocity = 3.00 %

E. coli bacterial cells are around 1 μm = 10^−6 m in length

Step 2: Calculate uncertainty in velocity

Δv = 0.03 * 8*10^-6 m/s =2.4 * 10^-7 m/s

Step 3: Calculate the uncertainty of the position of the bacterium

According to Heisenberg uncertainty principle,

Δx *Δp ≥ h/4π

Δx *mΔv ≥ h/4π

with Δx = TO BE DETERMINED

with m = 1.8 *10^-18 kg

with Δv = 2.4*10^-7

with h = constant of planck = 6.626 *10^-34

Δx ≥ 6.626*10^-34 / (4π*(1.8*10^-18)(2.4*10^-7))

Δx ≥ 1.22 *10^-10m

6 0

3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

3 years ago

How many grams of potassium chlorate must be heated to produce 30.0g of oxygen? socratic.org

DanielleElmas [232]

Answer:

76.56g

Explanation:

Firstly, to do this we need a correct and balanced equation for the decomposition of potassium chlorate.

2KClO3 —-> 2KCl + 3O2

From the balanced equation, we can see that 2 moles of potassium chlorate yielded 3 moles of oxygen gas

We need to know the actual number of moles of oxygen gas produced. To do this, we divide the mass of the oxygen gas by its molar mass. Its molar mass is 32g/mol

The number of moles is thus 30/32 = 0.9375 moles

Now we can calculate the number of moles of potassium chlorate decomposed.

We simply do this by (0.9375 * 2)/3 = 0.625 moles

Now to get the number of grammes of potassium chlorate decomposed, we simply multiply this number of moles by the molecular mass. The molecular mass of KClO3 is 39 + 35.5 + 3(16) = 122.5g/mol

The amount in grammes is thus 122.5 * 0.625 = 76.56g

5 0

3 years ago

Read 2 more answers