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3 years ago

Each incomplete reaction is a neutralization reaction. Match each equation with the salt formed.

2 answers:

6 0

Based on the incomplete neutralization reaction above, the corresponding salts that each reaction would form are as follows:
1. H2SO4 + 2NH4OH----(NH4)2SO4 with water as a byproduct
2. 2NaOH + H2CO3 ----Na2CO3 with water as a byproduct
3. HNO3 + KOH ----KNO3 with water as a byproduct

Serggg [28]3 years ago

6 0

Explanation:

Neutralization reaction Is chemical reaction between acid and base which gives water and salt as product.

$2H_2SO_4+2NH4OH\rightarrow 2(NH_4)_2SO_4+3H_2O$

The salt formed in reaction will be $(NH_4)_2SO_4$ .

$2NaOH+H_2CO_3\rightarrow Na_2CO_3+2H_2O$

The salt formed in reaction will be $Na_2CO_3$

$HNO_3+KOH\rightarrow KNO_3+H_2O$

The salt formed in reaction will be $KNO_3$ .

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How many moles would you need to have 7.85 x 10^22 molecules of ammonia (NH3).

RSB [31]

<h3>Answer:</h3>

0.130 moles NH₃

<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>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 7.85 × 10²² molecules NH₃

[Solve] moles NH₃

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 7.85 \cdot 10^{22} \ molecules \ NH_3(\frac{1 \ mol \ NH_3}{6.022 \cdot 10^{23} \ molecules \ NH_3})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 0.130355 \ moles \ NH_3$

<u>Step 4: Check</u>

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

0.130355 moles NH₃ ≈ 0.130 moles NH₃

3 0

3 years ago

If the reaction was reversed and we wanted to produce as much NaN3, in grams, as possible from 30.0 g of N2 and 20.0 g of Na, wh

Valentin [98]

Answer:

N₂

Explanation:

Sodium is a larger molecule with a much higher molecular weight. However, 20g of N₂ would be a smaller amount of molecules than 20g of sodium due to how there are multiple nitrogen molecules.

2/3 * given mass of N₂ = mass of N₃

N₃ (Azide ion) given mass = 20g

Na = 20g

Masses of chemicals are equal

Na = 22.990g/mol

20g/22.990g/mol = 0.8699mol of Na

N₃ = 20g

N₃ = N g/mol x 3

N = 14.007 g/mol

14.007 x 3 = 42.021 g/mol

N₃ = 42.021g/mol

20g/42.021g/mol = 0.4759 mol of N₃

Notice how there are fewer moles of the Azide ion than the Sodium.

0.4759 moles of NaN₃ is produced

combine molecular weights:

42.021 + 22.990 = 65.011 g/mol

multiply by amount of moles of the limiting reactant:

0.4759 mol *65.011 = 30.942 g

Also, here is the balanced equation:

3N₂ + 2Na = 2NaN₃

The results are the same as the balanced equation.

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3 years ago

A scientist and his friend walked into a bar. The Scientist said: "May I please get some H2O". the waitress asked his friend wha

Aleksandr [31]

Answer:

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3 years ago

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What is the definition of polar in Chemistry?

sweet [91]

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3 years ago

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Anna [14]

Answer:

Explanation:

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3 years ago