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4 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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Liquid nitrogen trichloride is heated in a 2.50−L closed reaction vessel until it decomposes completely to gaseous elements. The

Fofino [41]

Answer:

1. Partial pressure of N2 is 204.5 mmHg

2. Partial pressure of Cl2 is 613.5 mmHg

Explanation:

Step 1:

The equation for the reaction. This is given below:

NCl3 —> N2 (g) + Cl2 (g)

Step 2:

Balancing the equation.

NCl3 (l) —> N2 (g) + Cl2 (g)

The above equation is balanced as follow:

There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by putting 2 in front of NCl3 as shown below:

2NCl3 (l) —> N2 (g) + Cl2 (g)

There are 6 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of Cl2 as shown below:

2NCl3 (l) —> N2 (g) + 3Cl2 (g)

Now the equation is balanced.

Step 2:

Determination of the mole fraction of each gas.

From the balanced equation above, the resulting mixture of the gas contains:

Mole of N2 = 1

Mole of Cl2 = 3

Total mole = 4

Therefore, the mole fraction for each gas is:

Mole fraction of N2 = mole of N2/total mole

Mole fraction of N2 = 1/4

Mole fraction of Cl2 = mole of Cl2/total mole

Mole fraction of Cl2 = 3/4

Step 3:

Determination of the partial pressure of N2.

Partial pressure = mole fraction x total pressure

Total pressure = 818 mmHg

Mole fraction of N2 = 1/4

Partial pressure of N2 = 1/4 x 818

Partial pressure of N2 = 204.5 mmHg

Step 4:

Determination of the partial pressure of Cl2.

Partial pressure = mole fraction x total pressure

Total pressure = 818 mmHg

Mole fraction of Cl2 = 3/4

Partial pressure of Cl2 = 3/4 x 818

Partial pressure of Cl2 = 613.5 mmHg

7 0

3 years ago

Read 2 more answers

3. Ethanol has a density of 0.800 g/mL. a. What is the mass of 225 mL of ethanol?

dedylja [7]

Since the ethanol has a density of 0.800 g/mL, the mass of the ethanol is 180 grams.

Given the following data:

Volume of ethanol = 225 mL

Density of lead ball = 0.800 g/mL.

To find the mass of the ethanol;

Density can be defined as mass all over the volume of an object.

Mathematically, the density of a substance is given by the formula;

$Density = \frac{Mass}{Volume}$

Making mass the subject of formula, we have;

$Mass = Density \times Volume$

Substituting the given parameters into the formula, we have;

Mass of ethanol = 180 grams.

3 0

3 years ago

12 POINTS!! Which step would help a student find the molecular formula of a compound from the empirical formula?

Novosadov [1.4K]

Your answer would be D.

5 0

3 years ago

What makes a balloon electrically charged?

Natalija [7]

C. Rubbing the balloon against your hair

3 0

2 years ago

Read 2 more answers

How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm?

meriva

Hello!

Data:

P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)

Apply the data to the Clapeyron equation (ideal gas equation), see:
$P*V = n*R*T$

$1*18.5 = n*0.082*300$

$18.5 = 24.6n$

$24.6n = 18.5$

$n = \dfrac{18.5}{24.6}$

$\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark$

Note: If the feedback is to be considered, the closest response is 0.751 mol Nacl

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I hope this helps. =)

3 0

3 years ago