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Mashutka [201]
3 years ago
5

Which of the following acid-naming rules is INCORRECT 

Chemistry
2 answers:
oksano4ka [1.4K]3 years ago
7 0

<u>Answer:</u> The incorrect statement is all acid names end with -ic.

<u>Explanation:</u>

Acids are defined as the chemical species which donate hydrogen ions when dissolved in water.

HA(aq.)\rightarrow H^+(aq.)+A^-(aq.)

The nomenclature of the acids that are derived from the anions attached to hydrogen ion:

  • If the name of the anion attached ends with -ate, then the prefix of the acid is -ic. <u>For Example:</u> H_3PO_4, the anion is phosphate and the name of the acid is phosphoric acid.
  • If the name of the anion attached ends with -ite, then the prefix of the acid is -ous. <u>For Example:</u> H_2SO_3, the anion is sulfite and the name of the acid is sulfurous acid.
  • If the name of the anion attached ends with -ide, then the prefix of the acid is -ic. <u>For Example:</u> HCl, the anion is chloride and the name of the acid is hydrochloric acid.

Hence, the incorrect statement is all acid names end with -ic.

ad-work [718]3 years ago
6 0
<span>The answer to this question would be: A. all acid names end with "-ic"

Not all acid have "-id" suffix for their name. The example of this was already mentioned in the question as the sulfite ion will be called sulfurous acid. The "-ous" name is given to acid with lower oxidation state. Sulfuric acid will be the H2SO4 which have higher oxidation state.</span>
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Balance the following chemical equation by providing the correct coefficients fe+h2so4 fe(so4)3 + h2
Dmitriy789 [7]

Answer:

2Fe + 3H2SO4 + Fe2(SO4)3+ 3H2

Explanation:

1. Fe (SO4) 3 is an incorrectly written formula because iron is trivalent as we can see by this three ahead of SO4. SO4 is divalent always.

2. since (SO4) is 3, this three shows us that there must be 3 in the reactants as well.

so now there is 3H2SO4

3. Since we have added 3 to one hydrogen we must add another. So now it's 3H2

4. and finally iron. In Fe2 (SO4) 3 we see this 2 in front of Fe which means it goes 2Fe.

3 0
2 years ago
7. If all of the forces on an object are balanced, the object?
dolphi86 [110]

Answer:

Is balanced

Explanation:

4 0
3 years ago
A 15.00 % by mass solution of lactose (C 12H 22O 11, 342.30 g/mol) in water has a density of 1.0602 g/mL at 20°C. What is the mo
Alik [6]

Answer:

22.82M

Explanation:

342.3g/mol is présent in 1000

what about in 15??

( 342.3g/mol × 1000 ) ÷ 15

3 0
2 years ago
Consider a situation in which 211 g
Stella [2.4K]

Answer:

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂       →     2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂ :

Number of moles = mass/ molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

                       O₂         :         P₂O₅

                        5          :           2

                        7.5       :        2/5×7.5 = 3.00

                       P₄          :         P₂O₅

                        1           :           2

                       1.7         :       2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

3 0
3 years ago
Which of the following gives the molarity of a 17.0% by mass solution of sodium acetate, CH 3COONa (molar mass = 82.0 g/mol) in
weeeeeb [17]

Answer:

The molarity of this solution is 2.26 M (option D)

Explanation:

Step 1: Data given

Mass % = 17%

Molar mass of CH3COONa = 82.0 g/mol

Density of the solution = 1.09 g/mL

Step 2:

Assume the mass of the solution is 1.00 gram

⇒ 17.0 % CH3COONa = 0.17 grams

⇒ 83.0 % H2O = 0.83 grams

Step 3:

Density = 1.09 g/mL

Volume of the solution = total mass / density of solution

Volume of solution : 1.00 grams / 1.09 g/mL

Volume of the solution = 0.917 mL = 0.000917 L

Step 4: Calculate  number of moles of CH3COONa

Number of moles = Mass / molar mass

Number of moles CH3COONa = 0.170 grams / 82.0 g/mol

Number of moles = CH3COONa = 0.00207 moles

Step 5: Calculate molarity

Molarity = moles / volume

Molarity solution =   0.00207 / 0.000917 L

Molarity solution =  2.26 mol/L = 2.26 M

The molarity of this solution is 2.26 M (option D)

6 0
3 years ago
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