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3 years ago

Write the product concentration expression for the decomposition of ammonia to hydrogen and nitrogen.

Chemistry

1 answer:

stich3 [128]3 years ago

4 0

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The titration of a 20.0-mLmL sample of an H2SO4H2SO4 solution of unknown concentration requires 22.87 mLmL of a 0.158 M KOHM KOH

kkurt [141]

Answer:

0.0905 M

Explanation:

Let's consider the neutralization reaction between H2SO4 and KOH.

H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O

22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:

0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol

The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol

1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:

M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M

3 0

3 years ago

Discuss the reason for the presence of large number of organic compounds

jeyben [28]

Answer:

Carbon forms the large numbers of compound due to the following reasons

4 0

3 years ago

one reaction that produces hydrogen gas can be represented by the unbalanced chemical equation Mg(s)+HCI(aq) -> MgCI(aq)+H2(g

Sonbull [250]

<h3>Answer:</h3>

128 g HCl

<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Reaction Mole Ratios
Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

↓

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

[Given] 3.25 mol Mg

[Solve] x g HCl

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Mg → 2 mol HCl

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

<u>Step 3: Stoich</u>

[S - DA] Set up: $\displaystyle 3.25 \ mol \ Mg(\frac{2 \ mol \ HCl}{2 \ mol \ Mg})(\frac{36.46 \ g \ HCl}{1 \ mol \ HCl})$
[S - DA] Multiply/Divide [Cancel out units]: $\displaystyle 127.61 \ g \ HCl$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

127.61 g HCl ≈ 128 g HCl

3 0

3 years ago

A 1.00 L of a solution is prepared by dissolving 125.6 g of NaF in it. What would be the molarity of this solution?

DedPeter [7]

Answer:

2.99 M

Explanation:

In order to solve this problem we need to keep in mind the definition of molarity:

Molarity = moles of solute / liters of solution

In order to calculate the moles of solute, we <u>convert 125.6 g of NaF into moles</u> using its <em>molar mass</em>:

125.6 g NaF ÷ 42 g/mol = 2.99 mol NaF

As the volume is already given, we can proceed to <em>calculate the molarity</em>:

Molarity = 2.99 mol / 1.00 L = 2.99 M

4 0

3 years ago

Which of the following is a balanced chemical equation?

True [87]

Answer:

equation number 3 is balanced.

hope it helps ☺️!

8 0

3 years ago