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Nonamiya [84]
3 years ago
15

So to find a base by its formula, does it end in COO or COOH? Can a base be both of those endings?

Chemistry
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

To find a base by its formula, does it end in COO-

A base never ens with COOH

Explanation:

According to Bronsted Acid - Base theory : When an acid and base reacts with each other , the acid form its conjugate base and base will form its conjugate acid because of exchange of proton .

If HA = acid and B = base react each other then they generate their conjugate Base (A-) and acid (HB+) respectively .

The equation is represented as :

HA+B\rightleftharpoons A^{-}+HB^{+}

Here A- is conjugate base of HA acid

HB+ is conjugate acid of base B

COOH group is the carboxylic acid group = acid

Suppose A carboxylic acid RCOOH is added in water(H2O)

RCOOH + H_{2}O\rightleftharpoons RCOO^{-}+H^{+}

Here

RCOOH = Acid and its conjugate -base is RCOO-

Hence A Base end with formula COO-

COOH is always acid

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4 years ago
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Which statement describes the bonds in iron sulfate, FeSO4?
Feliz [49]

___

Regarding the bonds in FesO₄, Fe and S have an ionic bond, while S and O have covalent bonds.

Elements form bonds to increase their stability. The main types of bonds are:

  • Metallic bonds: they are formed between metals and the electrons are in a delocalized cloud.
  • Ionic bonds: they are formed between metals (lose electrons) and nonmetals (gain electrons)
  • Covalent bonds: they are formed between nonmetals, which share electrons.

Regarding the bonds in FesO₄:

  • Fe is a metal and S a nonmetal, thus they will form ionic bonds.
  • S and O are both nonmetals, thus they will form covalent bonds.

Regarding the bonds in FesO₄, Fe and S have an ionic bond, while S and O have covalent bonds.

Learn more: brainly.com/question/23882847

5 0
2 years ago
Which units are used to measure both velocity and speed? Select three options
Ede4ka [16]

Answer and Explanation:

The options aren't listed in your question, but here are some units that are regularly and normally used (in the classroom and in the outside world):

(The SI unit of distance and displacement is the meter. The SI unit of time is the second.)

<u>Meters per Second (m/s)</u>

kilometers per hour (km/hr)

kilometers per second (km/sec)

To find the average speed, you do distance divided by time.

To find the average velocity, you do the final position minus the initial position, divided by the final time minus the initial time.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

<em><u></u></em>

<em><u>I hope this helps!</u></em>

8 0
3 years ago
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.

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