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3 years ago

Describe the chemical reaction based on the chemical equation below. Also, explain whether the equation is balanced.

Chemistry

2 answers:

Allushta [10]3 years ago

5 0

Answer:

Chemical equation: NH₃(g) + O₂(g) → NO(g) + H₂O(l).

Oxidation reaction: N⁻³ → N⁺² + 5e⁻ /×4.

Reduction reaction: O₂⁰ + 4e⁻ → 2O⁻² /×5.

Oxidation reaction: 4N⁻³ → 4N⁺² + 20e⁻.

Reduction reaction: 5O₂⁰ + 20e⁻ → 10O⁻² /×5.

Balanced chemical equation: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l).

Ammonia (NH₃) reacts with oxygen (O₂) and form nitrogen(II) oxide (NO) and water (H₂O). Catalyst is this reaction is platinum (Pt).

This is combustion reaction.

katrin2010 [14]3 years ago

4 0

Answer:

Ammonia gas reacts with oxygen gas.

Nitric oxide gas and liquid water are produced.

Platinum is used as a catalyst.

The equation is unbalanced because the number of hydrogen atoms is not the same on both sides of the equation.

Explanation:

exact edg answers

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Find the empirical formula of the following compounds:

Aneli [31]

The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.

<h3>What is empirical formula?</h3>

The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.

<h3>How to find the empirical formula?</h3>

Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.

Moles phosphorus = 0.903 g phosphorus $\frac{mol phosphorus}{ 30.97 g phosphorus}$ = 0.0293 mol

Moles bromine 6.99 g bromine $\frac{mol bromine}{79.90 g bromine}$ =0.0875 mol

The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3

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1 year ago

Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m

Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

$molarity = moles of solute ÷ liters of solution$

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

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