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harkovskaia [24]
3 years ago
9

Atoms that are sp2 hybridized form how many pi bonds?

Chemistry
1 answer:
vova2212 [387]3 years ago
3 0
Sp2 hybridization forms 1 sigma bond and 1 pi bond.
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Ammonium nitrate dissociates in water according to the following equation:
Viktor [21]

Answer:

Explanation:

NH₄NO₃ = NH₄⁺ +NO₃⁻

heat released  by water = msΔ T

m is mass , s is specific heat and ΔT is fall in temperature

= 50  x 4.18 x ( 22 - 16.5 )  ( mass of 50 mL is 50 g )

= 1149.5 J .

This heat will be absorbed by the reaction above .

q for the reaction = + 1149.5 J

2 )

molecular weight of NH₄NO₃ = 80

No of moles reacted = 5/80 = 1 / 16 moles.

3 )  

5 g absorbs 1149.5 J

80 g absorbs 1149.5 x 16 J

= 18392 J

= 18.392 kJ.

= + 18.392 kJ

ΔH =  18.392 kJ / mol

6 0
3 years ago
The bright-line spectrum of an element in the gaseous phase is produced as
sesenic [268]
Electrons move from higher energy states to lower energy states Hope this helped
3 0
3 years ago
If it takes 5 hours to produce 8.0 kg of alcohol, how any days will it take to consume 1000 kg of glucose?
Ipatiy [6.2K]

It takes 21.3 days

<h3>Further explanation</h3>

Given

5 hr = 8 kg Alcohol

Required

Days to consume 1000 kg of glucose

Solution

Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,

C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂

mol ethanol :

\tt \dfrac{5000~g}{46~g/mol}=108.7~moles

moles of glucose to produce 108.7 moles ethanol :

\tt \dfrac{1}{2}\times 108.7=54.35

54.35 moles = 5 hours

moles of 1000 kg of glucose :

\tt \dfrac{10^6~g}{180~g/mol}=5555.5~moles

So for 5555.5 moles, it takes :

\tt \dfrac{5555.5}{54.35}\times 5~hours=511.085~h=21.3~days

7 0
3 years ago
You have a pound of feathers and a pound of lead:
mafiozo [28]
Feathers have more density and less have more mass, feather is super light and a POUND of led isnt
6 0
3 years ago
Read 2 more answers
A solution is prepared by dissolving 0.56 g of benzoic acid (C6H5CO2H, Ka ???? 6.4 ???? 10????5) in enough water to make 1.0 L o
garri49 [273]

Answer:

[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22

Explanation:

Data: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵

M (molar mass) of BA (Benzoic Acid) = 122 g/mol

Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M

We should consider the equation once it reaches the equilibrium:

C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺

  C - x                      x              x

And, for the Kₐ:

Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x²/(C-x) , where C = 4.59x10⁻³

Then: x² + Kₐx - KₐC = 0

x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0

Resolving this cuadratic equation (remember to use Baskara equation), we obtain:

x = 6.083x10⁻⁴ M

Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M

[C₆H₅COOH] = C - x = 3.98x10⁻³ M

pH = -Log [H⁺] = 3.22

7 0
3 years ago
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