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2 years ago

How many mL of .450 M HCL is needed to neutralize 30mL of .150 M Ba(OH)2

Chemistry

1 answer:

Aleks04 [339]2 years ago

7 0

You need .556M HCL to neutralize that

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Ammonia (NH3) reacts with sulfuric acid to form ammonium sulfate. How many grams of ammonium sulfate are obtained from 8.73 g of

Gwar [14]

I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra

4 0

3 years ago

Pollution comes from many small sources.

Digiron [165]

Answer: True

Hope this helps

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3 years ago

What is the basic of an organic molecule

Deffense [45]

Organic molecules are the molecules of life and are built around chains of carbon atoms that are often quite long. There are four main groups of organic moleculesthat combine to build cells and their parts: carbohydrates, proteins, lipids, and nucleic acids.What is the basic of an organic molecule

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3 years ago

Read 2 more answers

6. A box measures 11.25 inches in length, 8.1 inches in width and 6.85 inches in height. What is the

denis23 [38]

Answer:

I'd say 624.2^3 inches.

Explanation:

3 0

2 years ago

A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3

lubasha [3.4K]

Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio [A⁻]/HA].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2 can be determined.

So,

4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA] = 4.63 - 4.20 = log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 = [A⁻]/[HA] = 2.69

[A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

Molarity is moles per liter of solution, so we can calculate how many moles of C6H5CO2⁻ the student needs to dissolve in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L = 0.34 mol C6H5CO2⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0

3 years ago