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3 years ago

Moles of phosphorous in 15-35-15 fertilizer in 10g

1 answer:

6 0

The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15% Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this fertilizer, to get the number of moles of phosphorus, you multiply the mass by 35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the calculated mass of phosphorous by its molar mass which is 30.97 g/mol. Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.

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Nitroglycerin is a dangerous powerful explosive that violently decomposes when it is shaken or dropped. The Swedish chemist Alfr

Ganezh [65]

Answer:

a. $4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas $N_2$ , oxygen gas $O_2$ , water vapor $H_2O$ and carbon dioxide $CO_2$ . Let's write the decomposition of nitroglycerin into these 4 components:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)$

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

$C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by $\frac{3}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)$

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by $\frac{5}{2}$ :

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

$\frac{5}{2} + 6 = 8.5$

This leaves $9 - 8.5 = 0.5 = \frac{1}{2}$ of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

$C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)$

To make it look neater without fractional coefficients, multiply both sides by 4:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

$V_{CO_2} = 41.0 L$

$T = -14.0^oC + 273.15 K = 259.15 K$

$p = 1 atm$

Firstly, we may find moles of carbon dioxide produced using the ideal gas law $pV = nRT$ .

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

$n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol$

According to the stoichiometry of the balanced chemical equation:

$4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)$

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

$\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol$

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

$m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g$

3 0

3 years ago

Nitric acid (HNO3) is a strong acid that is completely ionized in aqueous solutions of concentrations ranging from 1% to 10% (1.

Alborosie

Given:

Concentration of HNO3 = 7.50 M

% dissociation of HNO3 = 33%

To determine:

The Ka of HNO3

Explanation:

Based on the given data

[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M

The dissociation equilibrium is-

HNO3 ↔ H+ + NO3-

I 7.50 0 0

C -2.48 +2.48 +2.48

E 5.02 2.48 2.48

Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23

Ans: Ka for HNO3 = 1.23

6 0

3 years ago

What is the mass of an element with 78 protons and 78 neutrons

Reika [66]

156. Atomic mass if the number of protons and neutrons within an atom.

4 0

3 years ago

Read 2 more answers

A gaseous mixture contains 5.0 moles of nitrogen and 10.0 moles of helium . What is the total pressure if the temperature is 25

jolli1 [7]

Answer:

A) 122 atm

Explanation:

PV = nRT

Solve for P --> P = nRT/V

n = 10.0 mol + 5.0 mol = 15.0 mol

R = 0.08206 L atm / mol K

T = 25 + 273 = 298 K

V = 3.0

P = (15.0)(0.08206)(298) / (3.0) = 122 atm

6 0

3 years ago

Which of these solutions are basic at 25 Â°C? Solution A: [OHâˆ’]=3.13Ã—10âˆ’7 M Solution C: [H3O+]=0.000747 M Solution B: [H3O+

solong [7]

Answer:

Are basic:

[OH⁻] = 3.13x10⁻⁷M and [H₃O⁺] = 9.55x10⁻⁹M

Explanation:

A solution is basic when pH = - log [H₃O⁺] is higher than 7.

It is possible to convert [OH⁻] to [H₃O⁺] using:

[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]

a. [OH⁻] = 3.13x10⁻⁷M

[H₃O⁺] = 1x10⁻¹⁴ / [3.13x10⁻⁷M]

[H₃O⁺] = 3.19x10⁻⁸M

pH = - log [H₃O⁺] = 7.50

[OH⁻] = 3.13x10⁻⁷M is basic

b. pH = -log [H₃O⁺] = - log 0.000747M = 3.13.

This solution is not basic

c. [H₃O⁺] = 9.55x10⁻⁹M

pH = 8.02

This solution is also basic.

8 0

3 years ago