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3 years ago

When Phosphoric Acid behaves according to arrhenius theory in water, what are the two products formed by the first proton

Chemistry

1 answer:

bagirrra123 [75]3 years ago

5 0

Answer:

Phosphoric acid is a triprotic acid. This means that it can dissociate in water up to three times, each time releasing a proton into the water as shown in the following reactions:

H3PO4 (s) + H2O (l) is in equilibrium with H3O + (aq) + H2PO4− (aq)

H2PO4− (aq) + H2O (l) is in equilibrium with H3O + (aq) + HPO42− (aq)

HPO42− (aq) + H2O (l) is in equilibrium with H3O + (aq) + PO43− (aq)

Explanation:

Phosphoric acid having contact with water, dissociating from a proton up to three times, that is why the three possible reactions are determined above.

This acid is an acid that belongs to oxo acids and its formula is H3PO4

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Based on location in the periodic table, fluorine (F) has chemical properties that are most similar to iodine.

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2 years ago

If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did

user100 [1]

Answer:

4.5 L water we have in litres (L).

Explanation:

$Q=m\times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

Plugging in the values

$\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$$

So,

Volume of water = mass/density

$\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$$

=4.5 L (Answer)

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