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Furkat [3]
4 years ago
8

How many acids are there

Chemistry
2 answers:
Galina-37 [17]4 years ago
6 0
<h2>There are 8 strong acids.</h2>

Explanation:

  • There are 8 strong acids known.
  • The other acids known are weak.

 The name of 8 strong acids are as follows:

  1. Hydrochloric acid
  2. Sulfuric acid
  3. Nitric acid
  4. Hydrobromic acid
  5. Hydroiodic acid
  6. Perchloric acid
  7. Chloric acid
  8. Periodic acid

Some of the weak acids are as follows-

        Citric acid, acetic acid, lactic acid, phosphoric acid, nitrous acid, hydrofluoric acid.

Sonja [21]4 years ago
4 0
There are at least 7 strong acids
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11Alexandr11 [23.1K]
The answer would be D, all the above
5 0
3 years ago
Read 2 more answers
Realizar una oracion empleando los verbos en infinitivo que estan anteriormente (acompañar, bailar, dormir, convivir, proteger)
gladu [14]

Answer:

(Léase explicación)

Explanation:

1) El presidente va a acompañar a la ciudadanía en tan histórica jornada.

2) No me gusta bailar en solitario.

3) Me gusta dormir después de degustar un chocolate caliente con malvavisco.

4) El Coronavirus ha demostrado que algunas familias no saben convivir entre sí.

5) Voy a proteger al débil frente a los embates del crimen organizado.

7 0
4 years ago
Consider the following reaction:2NOBr(g) 2NO(g) + Br2(g)If 0.412 moles of NOBr(g), 0.678 moles of NO, and 0.224 moles of Br2 are
Vaselesa [24]

Answer: K_c=0.0587M

Explanation: The given chemical reaction is:

2NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)

Equilibrium constant (Kc) in general is written as:

K_c=\frac{[products]}{[reactants]}

Note:- Coefficients are written as their powers

So, the Kc expression for the above reaction will be:

K_c=\frac{[NO]^2[Br_2]}{[NOBr]^2}

Equilibrium moles are given for all of them. Let's divide the moles by given liters to get the concentrations.

[NOBr]=\frac{0.412mol}{10.3L}   = 0.040 M

[NO]=\frac{0.678mol}{10.3L}   = 0.0658 M

[Br_2]=\frac{0.224mol}{10.3L}   = 0.0217 M

Plug in the values in the equilibrium expression to calculate Kc.

K_c=\frac{(0.0658M)^2(0.0217M)}{(0.040M)^2}

K_c=0.0587M

4 0
3 years ago
The reactant atoms are copper, hydrogen, nitrogen, and oxygen. Which reactant atom was oxidized in the reaction?
SCORPION-xisa [38]

Answer:

C. copper.

Explanation:

  • The atom which loses electrons (its oxidation sate be more positive) is the atom that is oxidized.
  • While, the atom which gains electrons (its oxidation sate be more negative) is the atom that is reduced.

  • Nitrogen:

It is oxidation sate is changed from (+5) in the reactants (NO₃⁻) to (+4) in the products (NO₂). N gains 1 electron

So, it is reduced.

  • Oxygen:

It is oxidation sate is the same (-2) in the reactants (NO₃⁻) and (-2) in the products (NO₂).

<em>So, it is neither be oxidized nor reduced.</em>

<em></em>

  • Copper:

It is oxidation sate is changed (0) in the reactants (Cu) to (+2) in the products (Cu²⁺). Cu loses 2 electrons.

<em>So, it is oxidized.</em>

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  • Hydrogen:

It is oxidation sate is the same (+1) in the reactants (H⁺) and (+1) in the products (H₂O).

<em>So, it is neither be oxidized nor reduced.</em>

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5 0
4 years ago
The concentration of hydrogen peroxide solution can be determined by
max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

8 0
3 years ago
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