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o-na [289]
3 years ago
14

The value of ΔH° for the reaction below is -6535 kJ. ________ kJ of heat are released in the combustion of 16.0 g of C6H6 (l)?

Chemistry
1 answer:
Umnica [9.8K]3 years ago
3 0

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

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Which statement best describes how the temperature of the ocean's surface water varies \
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3 years ago
The density of a 20.3 m ch3oh (methanol) solution is 0.858 g/ml. what is the molality of this solution? h2o is the solvent.
aliina [53]

The difference of molarity and molality is in the units:

Molarity is expressed in mol solute / L solution;

<span> while Molality is expressed in mol solute / kg solvent</span>

Since no other data is given, let us assume that the contribution of methanol on the total volume is negligible. So that,

Molality = (20.3 mol / L) (1 L / 0.858 kg)

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6 0
4 years ago
Phosphoric acid is a triprotic acid ( K a1 = 6.9 × 10 − 3 Ka1=6.9×10−3, K a2 = 6.2 × 10 − 8 Ka2=6.2×10−8, and K a3 = 4.8 × 10 −
irina1246 [14]

<u>Answer:</u> To calculate the pH of the buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a2

<u>Explanation:</u>

Phosphoric acid is a triprotic acid and it will undergo three dissociation reaction each having their respective dissociation constants.

The chemical equation for the first dissociation reaction follows:

H_3PO_4\rightleftharpoons H_2PO_4^-+H^+;K_a1=6.9\times 10^{-3}

The chemical equation for the second dissociation reaction follows:

H_2PO_4^-\rightleftharpoons HPO_4^{2-}+H^+;K_a2=6.2\times 10^{-8}

The chemical equation for the third dissociation reaction follows:

HPO_4^{2-}\rightleftharpoons PO_4^{3-}+H^+;K_a3=4.8\times 10^{-13}

To form a buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a of second dissociation process

To calculate the pK_a, we use the equation:

pK_a=-\log (K_a)\\\\pK_a=-\log(6.2\times 10^{-8})\\\\pK_a=7.21

To calculate the pH of buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a2+\log(\frac{[\text{conjugate base}]}{[\text{weak acid}]})

pH=pK_a2+\log(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]})

We are given:

pK_a2 = negative logarithm of second acid dissociation constant of phosphoric acid = 7.21

[HPO_4^{2-}] = concentration of conjugate base

[H_2PO_4^{-}] = concentration of weak acid

Hence, to calculate the pH of the buffer composed of H_2PO_4^-\text{ and }HPO_4^{-2}, we use the K_a2

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4 years ago
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