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3 years ago

8

Write the formula of the conjugate base of the Brønsted-Lowry acic, HF

1 answer:

alexdok [17]3 years ago

5 0

Answer:

Furthermore, because hydrofluoric acid, HF, loses a proton, H+1, to generate its conjugate in the reaction that is shown above, this reactant can be classified as a Brønsted-Lowry acid, and, consequently, the fluoride ion, F–1, is the conjugate base of this acid.

Explanation:

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How many lead atoms are in a cube of lead that has a volume of 2.18 cm3 if the density of lead is 11.3 g/cm3?

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3 years ago

Calculate ΔH∘f for NO(g) at 435 K, assuming that the heat capacities of reactants and products are constant over the temperature

weeeeeb [17]

Answer:

91383 J

Explanation:

The equation of the reaction can be represented as:

$\frac{1}{2} N_{2(g)}+\frac{1}{2} O_{2(g)}$ ------> $NO_{(g)}$

Given that:

The standard enthalpy of formation of NO(g) is 91.3 kJ⋅mol−1 at 298.15 K.

The equation below shown the reaction between the enthalpy of reaction at a particular temperature to another.

$\delta H^0__{R,T_2}$ = $\delta H^0__{R,T_1} } + \int\limits^{T_2}_{T_1} {\delta C_p(T')} \, dT'$

where:

$\delta H^0__{R}$ = enthalpy of reaction

${\delta C_p(T')}$ = the difference in the heat capacities of the products and the reactants.

∴

$\delta H^0__{R,435K}$ = $\delta H^0__{R,298.15K}$ $+$ $\int\limits^{435}_{298.15} {\delta C_p(T')} \, dT'$

= $1(91300 J.mol^{-1} ) +\int\limits^{435}_{298.15} [{(29.86)-\frac{1}{2}(29.38)-\frac{1}{2}29.13}]J.K^{-1}.mol^{-1} \, dT'$

= 91300 J + (0.605 J.K⁻¹)(435-298.15)K

= 91382.79 J

$\delta H^0__{R,435K}$ ≅ 91383 J

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3 years ago

Write the ground-state electron configuration for the chloride ion.You may write either the full or condensed electron configura

dalvyx [7]

Answer:

$1s^22s^22p^63s^23p^6$

Explanation:

Chlorine is the element of group 17 and third period. The atomic number of chlorine is 17 and the symbol of the element is Cl.

The electronic configuration of the element chlorine is:-

$1s^22s^22p^63s^23p^5$

Chloride ion is formed when chlorine atom gain one more electron. So, the ground-state electron configuration for the chloride ion is:-

$1s^22s^22p^63s^23p^6$

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