Write the conventional, total ionic, and net ionic equations for Mg(s)+H2O(l)

Rank the following fertilizers in decreasing order of mass percentage of nitrogen:

charle [14.2K]

<h3>Answer:</h3>

NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

<h3>Soution:</h3>

In (NH₄)₂HPO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄ = 132.06 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age = 21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age = 28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age = 21.18 %

In KNO₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of KNO₃ = 101.10 g.mol⁻¹

Mass %age = Mass of N / M.Mass of KNO₃ × 100

Mass %age = 14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age = 13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄ = 115.03 g.mol⁻¹

Mass %age = Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age = 14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age = 12.17 %

In NH₃:

Mass of Nitrogen = N × 1 = 14 × 1 = 14 g.mol⁻¹

Molar Mass of NH₃ = 132.14 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₃ × 100

Mass %age = 14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age = 82.20 %

In NH₄NO₃:

Mass of Nitrogen = N × 2 = 14 × 2 = 28 g.mol⁻¹

Molar Mass of NH₄NO₃ = 80.04 g.mol⁻¹

Mass %age = Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age = 28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age = 34.98 %

5 0

3 years ago

Pleaseeeeeefgghhbbbbbbbbbb

ehidna [41]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>----------------------------------------------------------------------</h2><h2>ELECTROPLATING </h2>

Electroplating the plating one metal on to the another metal, It is mostly used for preventing corrosion by using copper or chromium or decorate the object by using gold or silver plating.

__________________________________________________________

<h3>Principle:</h3>

When electricity is passed the thin layer of metal is deposited on another metal and water molecule given out as a By-Product, Thus this process works on the principle of Hydrolysis.

<h2>_____________________________________</h2><h2>QUESTION:</h2>

A) Electrolysis

B) Chromium prevents corrosion and gives the fine shining touch to the objects.

C) The metal which is deposited to the object i.e. spoon will be connected to the positive electrode of a battery, Thus it is anode. The spoon at which electroplating is need to be done is connected to the negative electrode, thus the Spoon is cathode.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

5 0

3 years ago

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles$

b) moles of $I_2$

$\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles$

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

According to stoichiometry :

1 mole of $I_2$ require 1 mole of $H_2$

Thus 0.0787 moles of $l_2$ require= $\frac{1}{1}\times 0.0787=0.0787moles$ of $H_2$

Thus $l_2$ is the limiting reagent as it limits the formation of product and $H_2$ acts as the excess reagent. (10.0-0.0787)= 9.92 moles of $H_2$ are left unreacted.

Mass of $H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g$

Thus 19.9 g of $H_2$ remains unreacted.

5 0

3 years ago

Calculate the volume in liters of a potassium iodide solution that contains of potassium iodide . Be sure your answer has the co

Anna11 [10]

Answer:

23.8 L

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.</em>

<em />

The molar mass of potassium iodide is 166.00 g/mol. The moles corresponding to 150 grams are:

150 g × (1 mol/166.00 g) = 0.904 mol

0.904 moles of potassium iodide are contained in an unknown volume of a 0.0380 mol/L potassium iodide solution. The volume is:

0.904 mol × (1 L/0.0380 mol) = 23.8 L

6 0

3 years ago

How much sodium bicarbonate to raise alkalinity in pool.

almond37 [142]

Answer:

A rule of thumb is that 1.5 lbs. of baking soda per 10,000 gallons of water will raise alkalinity by about 10 ppm. If your pool's pH is tested below 7.2, add 3-4 pounds of baking soda. If you're new to adding pool chemicals, start by adding only one-half or three-fourths of the recommended amount.

4 0

2 years ago