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Minchanka [31]
3 years ago
13

What mass of water would be produced if 1.5 g of ethanoic acid was reacted completely with sodium hydroxide (3 marks)

Chemistry
1 answer:
Zigmanuir [339]3 years ago
5 0

Mass of water : 0.45 g

<h3>Further explanation</h3>

Given

1.5 g ethanoic acid

Required

Mass of water

Solution

Reaction

<em>CH₃COOH + NaOH ⇒H₂O + CH₃COONa</em>

mol CH₃COOH(MW=60 g/mol) :

= mass : MW

= 1.5 : 60

= 0.025

mol H₂O from the equation :

= 0.025(mol ratio CH₃COOH : H₂O= 1 : 1)

Mass of water :

= mol x MW water

= 0.025 x 18 g/mol

= 0.45 g

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saw5 [17]

The periodic table is one of the most important tools in the history of chemistry. It describes the atomic properties of every known chemical element in a concise format, including the atomic number, atomic mass and relationships between the elements. Elements with similar chemical properties are arranged in columns in the periodic table.

The table thus is a quick reference as to what elements may behave the same chemically or which may have similar weights or atomic structures.

Hope this answer helps you

4 0
3 years ago
What is the molarity of a solution containing 8.9 g of NaOH in 550. mL of NaOH solution?
bulgar [2K]

Answer:

0.4 M

Explanation:

Molarity is defined as moles of solute, which in your case is sodium hydroxide,  

NaOH

, divided by liters of solution.

molarity

=

moles of solute

liters of solution

Notice that the problem provides you with the volume of the solution, but that the volume is expressed in milliliters,  

mL

.

Moreover, you don't have the number of moles of sodium hydroxide, you just have the mass in grams. So, your strategy here will be to

determine how many moles of sodium hydroxide you have in that many grams

convert the volume of the solution from milliliters to liters

So, to get the number of moles of solute, use sodium hydroxide's molar mass, which tells you what the mass of one mole of sodium hydroxide is.

7

g

⋅

1 mole NaOH

40.0

g

=

0.175 moles NaOH

The volume of the solution in liters will be

500

mL

⋅

1 L

1000

mL

=

0.5 L

Therefore, the molarity of the solution will be

c

=

n

V

c

=

0.175 moles

0.5 L

=

0.35 M

Rounded to one sig fig, the answer will be

c

=

0.4 M

Explanation:

6 0
3 years ago
Which of the following statements describes a physical property?
castortr0y [4]


The substance has a higher density than water


8 0
3 years ago
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What is atom?????????????
kiruha [24]

Answer:

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6 0
2 years ago
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Which of the following pairs lists a substance that can neutralize H2SO4 and the salt that would be produced from the reaction?
Bogdan [553]

The second option only.

  • LiOH, Li₂SO₄.
<h3>Explanation</h3>

A base neutralizes an acid when the two reacts to produce water and a salt.

Sulfuric acid H₂SO₄ is the acid here. There are more than one classes of bases that can neutralize H₂SO₄. Among the options, there are:

Metal hydroxides

  • Ca(OH)₂ and
  • LiOH.

Metal hydroxides react with sulfuric acid to produce water and the sulfate salt of the metal.

\text{Ca}(\text{OH})_{\bf 2}+\text{H}_2\text{SO}_4 \to \textbf{Ca}\textbf{SO}_{\bf 4} +{\bf 2}\;\text{H}_2\text{O}.

The formula for calcium sulfate \text{CaSO}_4 in option A is spelled incorrectly. Why? The charge on each calcium \text{Ca}^{2+} is +2. The charge on each sulfate ion {\text{SO}_4}^{2-} is -2. Unlike \text{Li}^{+} ions, it takes only one \text{Ca}^{2+} ion to balance the charge on each {\text{SO}_4}^{2-} ion. As a result, \text{Ca}^{2+} and {\text{SO}_4}^{2-} ions in calcium sulfate exist on a 1:1 ratio.

2\;\text{LiOH} +\text{H}_2\text{SO}_4 \to \text{Li}_2\text{SO}_4 + 2\;\text{H}_2\text{O}.

Ammonia, NH₃

Ammonia NH₃ can also act as a base and neutralize acids. NH₃ exists as NH₄OH in water:

\text{NH}_3 + \text{H}_2\text{O} \to \textbf{NH}_{\bf 4}\text{OH}.

The ion {\text{NH}_4}^{+} acts like a metal cation. Similarly to the metal hydroxides, NH₃ (or NH₄OH) neutralizes H₂SO₄ to produce water and a salt:

2\;\textbf{NH}_{\bf 4}\text{OH}+ \text{H}_2\text{SO}_4 \to (\textbf{NH}_{\bf 4})_2\text{SO}_4+2\;\text{H}_2\text{O}.

The formula of the salt (NH₄)₂SO₄ in the fourth option spelled the ammonium ion incorrectly.

As part of the salt (NH₄)₂SO₄, the ammonium ion NH₄⁺ is one of the products of this reaction and can't neutralize H₂SO₄ any further.

7 0
3 years ago
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