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astraxan [27]
3 years ago
8

Help me with this questions please

Chemistry
1 answer:
mina [271]3 years ago
8 0
I will figure out the answer and i will letu know what else do u want to talk about .
do u want to talk about periods
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2 H2O2(aq) ----> 2 H2O(l) + O2(g) in the presence of I-(aq) is proposed to be: Step 1 (slow): H2O2 + I- -----> H2O + OI- S
andrezito [222]

Answer:

Molecularity of the rate determining step = 2

Explanation:

Step 1 (slow): H₂O₂ + I⁻ -----> H₂O + OI⁻

Step 2 (fast): H₂O₂ + OI⁻ -----> H₂O + O₂ + I⁻

The rate determining step in a reaction mechanism is also considered as slowest step.

Slowest step is also considered its highest activation energy in energy profile diagram.

In this case intermediate  (IO⁻) is formed.

Step 1 considered as a slowest step.

So,  Rate = K [H₂O₂][I⁻]

  Molecularity = 2

6 0
3 years ago
______ is a chemical property of matter.
Reika [66]
I think it’s chemical reactivity
4 0
3 years ago
The atomic number of carbon is 6. its nucleus must contain:
Aloiza [94]
Must contain: 6 protons, 6 electrons and 12 neutrons.
5 0
3 years ago
The electron dot diagram for an atom of carbon has how many pair(s) of electrons and how many unpaired electrons
krok68 [10]

Answer:

The resultant structure is shown below. This structure contains four shared pairs of electrons, which are located on all four "sides" of carbon's electron dot structure. Each of these shared pairs was created by pairing one of carbon's unpaired electrons with an unpaired electron from chlorine.

Explanation:

4 0
3 years ago
Suppose that 25.0 mL of 0.10 M CH3COOH (aq) is titrated with 0.10 M NaOH (aq). What is the pH after the addition of 10.0 mL of 0
olya-2409 [2.1K]

Answer:

pH=-1.37

Explanation:

We are given that 25 mL of 0.10 M CH_3COOH is titrated with 0.10 M NaOH(aq).

We have to find the pH of solution

Volume of CH_3COOH=25mL=0.025 L

Volume of NaoH=0.01 L

Volume of solution =25 +10=35 mL=\frac{35}{1000}=0.035 L

Because 1 L=1000 mL

Molarity of NaOH=Concentration OH-=0.10M

Concentration of H+= Molarity of CH_3COOH=0.10 M

Number of moles of H+=Molarity multiply by volume of given acid

Number of moles of H+=0.10\times 0.025=0.0025 moles

Number of moles of OH^-=0.10\times 0.01=0.001mole

Number of moles of H+ remaining after adding 10 mL base = 0.0025-0.001=0.0015 moles

Concentration of H+=\frac{0.0015}{0.035}=4.28\times 10^{-2} m/L

pH=-log [H+]=-log [4.28\times 10^{-2}]=-log4.28+2 log 10=-0.631+2

pH=-1.37

6 0
3 years ago
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