what is the concentration in ppm of a solution which is prepared by dissolving in 15mg of nacl in 200ml water

Identify the limiting reactant in the reaction of nitrogen and hydrogen to form NH3 if 5.23 g of N2 and 5.52 g of H2 are combine

Marrrta [24]

Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

2) The amount (in grams) of excess reactant H₂ = 4.39 g.

Explanation:

Firstly, we should write the balanced equation of the reaction:

N₂ + 3H₂ → 2NH₃.

1) To determine the limiting reactant of the reaction:

From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.

This means that N₂ reacts with H₂ with a ratio of (1:3).

We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation: n = mass / molar mass.

The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

2) To determine the amount (in grams) of excess reactant of the reaction:

As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.

∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.

4 0

3 years ago

(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW

Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = 2.3x10⁻³M InF₃. Thus:

[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³

[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = 7.9x10⁻¹⁰

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, solubility is 6.31x10⁻⁶M

3 0

4 years ago

How many milliliters of 2.00 M H2SO4 will react with 28.0 g of NaOH?

77julia77 [94]

Answer:175⋅mL of the given sulfuric acid

Explanation:

5 0

3 years ago

The blank is the accepted form of gathering and reporting information within the science community.

Charra [1.4K]

Answer:

Scientific method.

Explanation:

Scientific method is the way taken to acquire scientific knowledge. It includes experiments, statistical analysis of existing data, and all kinds of observations of the world around us, while theoretical research is based on deriving certain theories about the world from basic principles, in a mathematical or logical way. The scientific method applies to both types of research, and emphasizes that scientific research is objective, that it can be verified by other scientists, and that knowledge is not acquired without context, but in a way that leads to a greater understanding of previous research and the world. we live in. To contribute to this, researchers are expected to clearly record both their findings and the methods they use to arrive at the results.

8 0

3 years ago

39. Analyze What subscripts would you most likely use if

Igoryamba

Based on their valencies, the subscripts of the ionic compounds formed will be:

1 and 1
2 and 1
1 and 2
2 and 2

<h3>What subscripts would you most likely use if the following substances formed an ionic compound?</h3>

A. An alkali metal and a halogen.

An alkali metal and a halogen both have valencies of one.

Therefore the subscripts would be 1.

B. An alkali metal and a non-metal from group 16.

An alkali metal has a valency of 1 and a group 16 non-metal has a valency of 2.

By exchange of valencies, the subscript would be 2 and 1.

C. An alkaline earth metal and a halogen.

An alkaline earth metal has a valency of 2 and a halogen has a valency of 1.

By exchange of valencies, the subscripts would be 1 and 2.

D. An alkaline earth metal and a non-metal from group 16.

An alkaline earth metal has a valency of 2 and a non-metal from group 16 has a valency of 2.

By exchange of valencies, the subscripts would be 2 and 2.

Therefore, the subscripts of the ionic compounds formed will be:

1 and 1
2 and 1
1 and 2
2 and 2

Learn more about about valency at: brainly.com/question/2284519

4 0

3 years ago