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ollegr [7]
3 years ago
14

Give the ΔH value for the decomposition of silver chloride as shown in the reaction 2AgCl(s)+127 kJ→2Ag(s)+Cl2(g).

Chemistry
1 answer:
choli [55]3 years ago
6 0

Answer: \Delta H=+127kJ

Explanation:

2AgCl(s)+127kJ\rightarrow 2Ag(s)+Cl_2(g)  

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and \Delta H for the reaction comes out to be negative.

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and \Delta H for the reaction comes out to be positive.

As the heat is absorbed the reactants , the reaction is endothermic and \Delta H for the given endothermic reaction will be +127 KJ

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When the submarine's density is equal to the density of the surrounding seawater, the submarine will maintain depth. If a 103200
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2.023 m^3 is the total displacement (volume) of the submarine.

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3 years ago
A chemistry graduate student is given 125.mL of a 1.00M benzoic acid HC6H5CO2 solution. Benzoic acid is a weak acid with =Ka×6.3
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Answer:

53.9 g

Explanation:

When talking about buffers is very common the problem involves the use of the Henderson Hasselbach formula:

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where  [A⁻] is the concentration of the conjugate base of the weak acid HA, and [HA] is the concentration of the weak acid.

We can calculate pKₐ from the given kₐ ( pKₐ = - log Kₐ ), and from there obtain the ratio  [A⁻]/HA].

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4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA] = - (-4.20 ) + log [A⁻]/[HA]

⇒ log [A⁻]/[HA]  = 4.63 - 4.20 =  log [A⁻]/[HA]

0.43 = log [A⁻]/[HA]

taking antilogs to both sides of this equation:

10^0.43 =  [A⁻]/[HA] = 2.69

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Molarity is moles per liter of solution, so we can calculate how many moles of  C6H5CO2⁻ the student needs to dissolve  in 125. mL ( 0.125 L ) of a 2.69 M solution:

( 2.69 mol C6H5CO2⁻ / 1L ) x 0.125 L  = 0.34 mol C6H5CO2⁻

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0.34 mol x 160.21 g/mol = 53.9 g

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