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Alisiya [41]
2 years ago
8

A. Indicate whether the following descriptions refer to electrolytes or to non-electrolytes.

Chemistry
1 answer:
Leokris [45]2 years ago
7 0

Answer:

1.non electrolytes

2.electrolytes

3.non electrolytes

4.electrolytes

5.electrolytes

6.electrolytes

Explanation:

electrolytes are substance in their molten form which can be decomposed by the passage of electricity exay includes acids,base and salts etc were as non electrolytes cannot be decomposed by the passage of electricity example alcohol,sugar organic solvents etc. Electrolytes in the body like potassium, sodium in their molten form helps to keep the electrical circuitry of the heart well

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What is the volume of a weather balloon that has 39 grams of helium with a density 0.017 g/mL?
olchik [2.2K]

Answer:

<h3>The answer is 235.29 mL</h3>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density}  \\

From the question.

density = 0.017 g/mL

mass = 4 g

We have

volume =  \frac{4}{0.017}  \\  = 235.2941176...

We have the final answer as

<h3>235.29 mL</h3>

Hope this helps you

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2 years ago
Can someone help me on 1 and 2
pav-90 [236]
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3 years ago
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Where are atoms located? (check all that apply)*
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3 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
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