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weqwewe [10]
3 years ago
6

What is the percent composition of C4H10? (4 points)

Chemistry
1 answer:
Marina CMI [18]3 years ago
6 0

Answer:82.7% C and 17.3% H

Explanation:

C4 = 12.0107g/mol x 4 = 48.0428 g/mol

H10 = 1.00794 g/mol x 10 = 10.0794 g/mol

Molar Mass of Butane, C4H10 =  48.0428 g/mol + 10.0794 g/mol= 58.1222 g/mol

percent composition of Carbon is  = ( mass of carbon contained in butane / molar mass of Butane) x 100

=(48.0428 /58.1222) x 100% = 0.8265 x 100

=82.65% =82.7% of Carbon.

Percent composition of Hydrogen  = (mass of Hydrogen contained in Butane / molar mass of Butane )x 100  

( 10.0794/58.122 ) x 100% =0.1734 x 100

= 17.3% OF Hydrogen.

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If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),
garri49 [273]

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

4 0
2 years ago
How many moles are there in 32.4 grams of Na
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Divide that my the molar mass which is 23 so 1.4087 g
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1)Which of the following changes would decrease the rate at which a solid solute dissolves in a liquid solvent? (3 points)
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Answer:

The first one is B, "Decreasing surface area."

Explanation:

This is because greater the surface area exposed, the more collisions that occur between the solvent and solute. I also just took the test myself and got it correct.

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Answer: how do we answer when there are no options??

Explanation:

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