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Maksim231197 [3]
4 years ago
9

The formula C2H4 can be described as

Chemistry
2 answers:
Mila [183]4 years ago
8 0
C2H4 is Ethylene.  Ethylene is a simple hydrocarbon alkaline.
Elden [556K]4 years ago
4 0
It's describe as Ethylene
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The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically. The following data was obtained: t/min 0
o-na [289]

Answer:

1) The order of the reaction is of FIRST ORDER

2)   Rate constant k = 5.667 × 10 ⁻⁴

Explanation:

From the given information:

The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.

liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.

The following data was obtained:

t/min                    0         10         20          30             40          ∞

conc B/(mol/L)    0       0.089    0.153     0.200       0.230    0.312

For  a first order reaction:

K = \dfrac{1}{t} \ In ( \dfrac{C_{\infty} - C_o}{C_{\infty} - C_t})

where :

K = proportionality  constant or the rate constant for the specific reaction rate

t = time of reaction

C_o = initial concentration at time t

C _{\infty} = final concentration at time t

C_t = concentration at time t

To start with the value of t when t = 10 mins

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 - 0}{0.312 - 0.089})

K_1 = \dfrac{1}{10} \ In ( \dfrac{0.312 }{0.223})

K_1 =0.03358 \  min^{-1}

K_1 \simeq 0.034 \  min^{-1}

When t = 20

K_2= \dfrac{1}{20} \ In ( \dfrac{0.312 - 0}{0.312 - 0.153})

K_2= 0.05 \times  \ In ( 1.9623)

K_2=0.03371 \ min^{-1}

K_2 \simeq 0.034 \ min^{-1}

When t = 30

K_3= \dfrac{1}{30} \ In ( \dfrac{0.312 - 0}{0.312 - 0.200})

K_3= 0.0333 \times  \ In ( \dfrac{0.312}{0.112})

K_3= 0.0333 \times  \ 1.0245

K_3 = 0.03412 \ min^{-1}

K_3 = 0.034 \ min^{-1}

When t = 40

K_4= \dfrac{1}{40} \ In ( \dfrac{0.312 - 0}{0.312 - 0.230})

K_4=0.025 \times  \ In ( \dfrac{0.312}{0.082})

K_4=0.025 \times  \ In ( 3.8048)

K_4=0.03340 \ min^{-1}

We can see that at the different time rates, the rate constant of k_1, k_2, k_3, and k_4 all have similar constant values

As such :

Rate constant k = 0.034 min⁻¹

Converting it to seconds ; we have :

60 seconds = 1 min

∴

0.034 min⁻¹ =(0.034/60) seconds

= 5.667 × 10 ⁻⁴ seconds

Rate constant k = 5.667 × 10 ⁻⁴

4 0
3 years ago
Use the chart to determine which type of bond is formed between potassium (K) and chlorine (Cl).
iris [78.8K]
Potassium and Chloride forms an ionic bond.
(K+) + (Cl-) = KCl

Potassium is under Group IA (Alkali Metal), wherein elements under this group can easily lose electrons.

Chlorine is under Group VII (Halogens), in which these elements can gain electrons easily.

The inner shell electrons on potassium will merge with the outer shell of electrons of chlorine to make potassium chloride.
6 0
4 years ago
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Calculate the pH of a buffer solution prepared by dissolving 0.20 mole of cyanic acid (HCNO) and 0.80 mole of sodium cyanate (Na
Afina-wow [57]

The pH of a buffer solution : 4.3

<h3>Further explanation</h3>

Given

0.2 mole HCNO

0.8 mole NaCNO

1 L solution

Required

pH buffer

Solution

Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.

\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}

valence according to the amount of salt anion  

Input the value :

\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3

7 0
3 years ago
How does the quantum theory of atoms explain the nature of the emission-line spectrum of hydrogen?
Oliga [24]

Answer:

Niels Bohr The electron in a Hydrogen atoms can only circle the nucleus in certain paths or orbits.

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3 years ago
14 L of a gas has a pressure of 6 atm. If the gas is compressed into 3.5 L what will the new pressure be?
hram777 [196]
Don’t really know what it is just need points end

5 0
3 years ago
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