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tester [92]
2 years ago
6

Write a balanced equation for the reactionbetween ammonia and sulfuric acid?​

Chemistry
1 answer:
Slav-nsk [51]2 years ago
8 0

Balanced equation for the reaction between ammonia and sulphuric acid is:

Explanation:

2NH3+H2SO4→(NH4)2SO4

I hope it'll help you...

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Which of the following prefixes would show up in the name for SF6? Hexa, penta, tetra, hepta, and mono
Mama L [17]

Answer:

Sulfur hexafluoride

Explanation:

5 0
2 years ago
Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
BabaBlast [244]

Answer:

The Ka is 3.74 *10^-4

Explanation:

Step 1: Data given

Mass of aspirin = 2.00 grams

Volume of water = 0.600 L

pH of solution = 2.61

Molar mass of acetylsalicylic acid = 180 g/mol

Step 2: The balanced equation

C9H8O4 + H2O ⇆ H3O+ + C9H7O4-

Step 3: Calculate moles acetylsalicylic acid

Moles acetylsalicylic acid = mass / molar mass

Moles acetylsalicylic acid = 2.00 grams / 180 g/mol

Moles acetylsalicylic acid = 0.0111 moles

Step 4: The initial concentration

[C9H8O4] = 0.0111 moles / 0.600 L = 0.0185 M

[H3O+] = 0M

[C9H8O4-] = 0M

Step 5: The concentration at the equilibrium

[C9H8O4] =0.0185 - X M

[H3O+] = XM

[C9H8O4-] = XM

Step 6: Calculate Ka

Ka = [H3O+][C9H7O4-] / [C9H8O4]

Ka = x² / (0.0185 - x)

pH = 2.61;  [H3O+] = 10^-pH = 10^-2.61 = 0.00245 = x

Ka = (0.00245)² / (0.0185 - 0.00245) = 3.74 * 10^-4

The Ka is 3.74 *10^-4

8 0
3 years ago
Identify the group or period that matches each description.
Mekhanik [1.2K]

Answer:

a.6th period

b.groupIIA

3 0
2 years ago
How can valence electrons be used to predict the type of compounds formed?
Alex17521 [72]
Valence electrons are the electrons in the outermost shell. Those are t<span>he electrons on an atom that can be gained or lost in a chemical reaction.
</span>Elements that are left on the periodic table <span> have relatively few </span>valence electrons<span>, and can form ions  more easily by losing their </span>valence electrons<span> to form positively charged cations.</span>
<span>Nonmetals are further to the right on the periodic table, so they gain electrons relatively easily and lose them with difficulty. </span>

7 0
3 years ago
1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium
lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

5 0
3 years ago
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