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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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You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with K
seropon [69]

Answer:

HA +  KOH  →  KA  +  H₂O

Explanation:

The unknown solid acid in water can release its proton as this:

HA  +  H₂O  →  H₃O⁺  +  A⁻

As we have the anion A⁻, when it bonded to the cation K⁺, salt can be generated, so the reaction of HA and KOH must be a neutralization one, where you form water and a salt

HA +  KOH  →  KA  +  H₂O

It is a neutralization reaction because H⁺ from the acid and OH⁻ from the base can be neutralized as water

7 0
3 years ago
What is the formula name for Co3N2
Ede4ka [16]

Answer:

Cobaltous Nitride,I think so anyway.......

6 0
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
lubasha [3.4K]

Answer:

53.9 g

Explanation:

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

pH = pKa + log [A⁻]/[HA]

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].

Since we know the concentration of HC6H5CO2 and the volume of solution, the moles and mass of KC6H5CO2  can be determined.

So,

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

 [A⁻]/ 1.00 M = 2.69 ⇒ [A⁻] = 2.69 M

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⁻

The mass will be obtained by multiplying 0.34 mol times molecular weight for KC6H5CO2 ( 160.21 g/mol ):

0.34 mol x 160.21 g/mol = 53.9 g

3 0
3 years ago
8.) If 396 g of Carbon Dioxide (CO2) are produced, what mass of Oxygen<br> (02) reacted? *
Brut [27]

Answer:

264g

Explanation:

C + O2 -> CO2

_g + _g -> 396g

396÷3=132

C (132g) + O2 (264g) -> CO2 (396g)

7 0
3 years ago
The solubility of kcl in ethanol is 0.25 g / 100 ml at 25 oc. how does this compare to the solubility of kcl in water?
Furkat [3]
Solubility data of a certain solute with a certain solvent is empirical. There are constant values for this at varying temperatures. For KCl in water at 25°C, the solubility is 35.7 g/100 mL of water. When you compare this with the solubility data of KCl with ethanol, this means that KCl is more soluble in water than in ethanol. This is true because KCl is an ionic salt which is very soluble in water.
8 0
3 years ago
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