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PSYCHO15rus [73]
3 years ago
11

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans

wer, only type in the numerical value with two significant figures. Do NOT include the unit.
Chemistry
1 answer:
Alik [6]3 years ago
4 0

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

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Maru [420]

When sodium amide i.e.NaNH_{2} reacts with water i.e. H_{2}O results in the formation of sodium hydroxide i.e. NaOH and ammonia NH_{3}.

The chemical reaction is given by:

NaNH_{2}+H_{2}O\rightarrow NH_{3}+NaOH

Now, when ammonia i.e.NH_{3} reacts with water results in the formation of ammonium hydroxide i.e. NH_{4}OH

The chemical reaction is given by:

NH_{3}+H_{2}O\rightarrow NH_4OH

Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).

The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.


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3 years ago
Convert mass to moles for both reactants. (round to 2 significant figures.)
Maslowich

Answer:

Explanation:

Given parameters:

Mass of CuCl₂  = 2.50g

Mass of Al  = 0.50g

Unknown:

Number of moles of CuCl₂ and Al  = ?

Solution:

To solve this problem, we must understand that the number of moles is a fundamental property used in stoichiometry calculations.

         Number of moles  = \frac{mass}{molar mass}

Molar mass of CuCl₂  = 63.6 + 2(35.5) = 134.5g/mole

Molar mass of Al  = 26.98g/mole

          Number of moles of  CuCl₂  = \frac{2.5}{134.5}  = 0.019moles

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3 years ago
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Problem page in each of the molecules drawn below one chemical bond is colored red. decide whether this bond is likely to be pol
Slav-nsk [51]

The question is incomplete, here is the complete question:

Problem page in each of the molecules drawn below one chemical bond is colored red. Decide whether this bond is likely to be polar or not. if the bond is likely to be polar, write down the chemical symbol for the atom which will have more negative charge.

The image is attached below.

<u>Answer:</u>

<u>For carbon dioxide molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.

<u>For water molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.

<u>Explanation:</u>

There are two types of covalent bonds:

  • <u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. <u>For Example:</u> CO_2,NO_2 etc..
  • <u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. <u>For Example:</u> N_2,O_2 etc..

<u>In carbon dioxide molecule:</u>

The given bond is present between C and O atom.

Electronegativity value of C = 2.5

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.5) = 1

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

<u>In water molecule:</u>

The given bond is present between H and O atom.

Electronegativity value of H = 2.1

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.1) = 1.4

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

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