All categories

2 years ago

Chlorine is more reactive than nitrogen why

1 answer:

4 0

But nitrogen is inert at room temperature whereas chlorine is highly reactive, the reason is: Nitrogen exists as N triple bond N at room temperature while chlorine exists as Cl single bond Cl at room temperature. Since the bond dissociation energy of triple bond is larger than that of single bond.

You might be interested in

The law of conservation of matter states that matter is neither created nor destroyed in a chemical reaction. Which of the follo

Artyom0805 [142]

Answer:

Explanation:

4 0

2 years ago

Read 2 more answers

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

Dmitry [639]

Answer:

7.5 g

Explanation:

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

<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

$4.9 g \times \frac{1mol}{98.00g} = 0.050mol$

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

$0.050mol \times \frac{149.09 g}{mol} = 7.5 g$

6 0

2 years ago

How many molecules of copper sulfate are needed to produce 5.0×1024 molecules of sodium sulfate?

lubasha [3.4K]

Answer:

5.0 × 10²⁴ molecules

Explanation:

Step 1: Write the balanced double displacement reaction

2 NaOH + CuSO₄ ⇒ Na₂SO₄ + Cu(OH)₂

Step 2: Calculate the moles corresponding to 5.0 × 10²⁴ molecules of Na₂SO₄

We will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

5.0 × 10²⁴ molecule × 1 mol/6.02 × 10²³ molecule = 8.3 mol

Step 3: Calculate the moles of CuSO₄ required to produce 8.3 moles of Na₂SO₄

The molar ratio of CuSO₄ to Na₂SO₄ is 1:1. The moles of CuSO₄ required are 1/1 × 8.3 mol = 8.3 mol.

Step 4: Calculate the molecules corresponding to 8.3 moles of CuSO₄

We will use Avogadro's number.

8.3 mol × 6.02 × 10²³ molecule/1 mol = 5.0 × 10²⁴ molecule

4 0

2 years ago

Which statement is TRUE about oxidation-reduction reactions? a. Every oxidation must be accompanied by a reduction. b. There are

Tomtit [17]

Answer:

Every oxidation must be accompanied by a reduction.

Explanation:

Oxidation and reduction are complementary processes. There can be no oxidation without reduction and vice versa. It is actually a given an take affair. A specie looses electrons which must be gained by another specie to complete the process. This explains why the selected option is the correct one.

8 0

2 years ago

The density of a 3.37M MgCl2 (FW = 95.21) is 1.25 g/mL. Calulate the molality, mass/mass percent, and mass/volume percent. So fa

Dafna1 [17]

Answer : The molality, mass/mass percent, and mass/volume percent are, 0.0381 mole/Kg, 25.67 % and 32.086 % respectively.

Solution : Given,

Density of solution = 1.25 g/ml

Molar mass of $MgCl_2$ (solute) = 95.21 g/mole

3.37 M magnesium chloride means that 3.37 gram of magnesium chloride is present in 1 liter of solution.

The volume of solution = 1 L = 1000 ml

Mass of $MgCl_2$ (solute) = 3.37 g

First we have to calculate the mass of solute.

$\text{Mass of }MgCl_2=\text{Moles of }MgCl_2\times \text{Molar mass of }MgCl_2$

$\text{Mass of }MgCl_2=3.37mole\times 95.21g/mole=320.86g$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=\text{Density of solution}\times \text{Volume of solution}=1.25g/ml\times 1000ml=1250g$

Mass of solvent = Mass of solution - Mass of solute = 1250 - 320.86 = 929.14 g

Now we have to calculate the molality of the solution.

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{3.37g\times 1000}{95.21g/mole\times 929.14g}=0.0381mole/Kg$