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3 years ago

What type of chemical reaction does this equation represent? H3PO3—>H3PO4 + PHz

Chemistry

1 answer:

Bogdan [553]3 years ago

6 0

Decomposition I believe

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Why is it wrong to assume that you are only "doing science" while you are doing experiments?

bazaltina [42]

Science is way to broad of a subject to define a specific thing you are doing. It is the equivalent of someone asking what book you are reading and you replying, "a novel." Your experiment could also have life-changing effects on someone, so to underrate it as "doing science" would simply be inaccurate.

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2 years ago

3. Methyl acetate is hydrolyzed at 25 oC in acidic environment. Aliquots of equal volume are removed and titrated with NaOH solu

avanturin [10]

Solution :

Time (sec) Volume of NaOH (mL)

339 26.23

1242 27.80

2745 29.70

4546 3.81

$\infty$ 39.81

Now the example of the first order kinetics w.r.t volumetric analysis is :

$k=\frac{2.303}{t} \log \left(\frac{v_{\infty}-v_0}{v_{\infty}- v_t}\right)$

Here, $v_{\infty}= \text{ volume at }\infty = 39.81$

$v_{t}= \text{ volume at time 't' } = 27.80$

$v_0$ = volume at time 0 = 0

Since the interval is not constant, we take the time interval as

$=\frac{903+1503+1801}{3}$

$=\frac{4207}{3}$

= 1402.3333

≈ 1402 seconds

$k=\frac{2.303}{1402} \log \left(\frac{39.81-0}{39.81-27.80}\right)$

$=(0.001643) \log \left(\frac{39.81}{10.01}\right)$

= 0.001643 x 0.52045

= 0.00082

$= 8.55 \times 10^{-4} \ sec^{-1}$

Therefore, the first order rate constant is k $= 8.55 \times 10^{-4} \ sec^{-1}$ .

5 0

2 years ago

Use your knowledge of valence electrons and how they affect bonds to figure out how carbon (Group 14) and oxygen (Group 16)

kenny6666 [7]

Answer: D

Explanation:

O=C=O

5 0

2 years ago

Green (Gg) Is crossed with a yellow pea plant (Gg )

Kobotan [32]

GG | Gg
-----------
Gg | gg

5 0

2 years ago

Read 2 more answers

Given 40 grams of magnesium. How many mole is this

Dahasolnce [82]

To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.

We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'

$\frac{40 g Mg}{1} x\frac{1 mole Mg}{24.305 g Mg}$

In this, we can see that the unit 'grams' will cancel out to leave us with moles.

In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.

3 0

3 years ago