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Kruka [31]
1 year ago
10

O2- is the Lewis ______ in the following reaction

Chemistry
1 answer:
matrenka [14]1 year ago
3 0

O^{2-} is the Lewis base in the following reaction:

SO_3(g) + O^{2-}(aq)   →  SO_4^{2-}(aq)

<h3>What is lewis base? </h3>

Lewis base is a species that donates an electron pair.

In the above-given chemical equation, the O^{2-}is the lewis base which is donating an electron pair to the central atom of SO_3 to form SO_4^{-2}.

Hence, O^{2-} is the Lewis base in the following reaction:

SO_3(g) + O^{2-}(aq)   →  SO_4^{2-}(aq)

Learn more about the lewis base here:

brainly.com/question/15570523

#SPJ1

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In this periodic table, which type of element is shown in green boxes?
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Answer: A

Explanation:

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Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxid
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sodium hexachloroplatinate(IV)- Na2[PtCl6]

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4 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

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According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

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5 0
3 years ago
Can someone help me name this hydrocarbon?
ser-zykov [4K]
2-ethyl-4,4 -dimethyl hex-1-ene.
3 0
3 years ago
What will be the new volume of a 250 mL sample of gas at 300 K and 1 atm if heated to 350 K at 1 atm?
AlekseyPX

The   new volume of a 250  Ml   sample of gas at 300k  and  1atm  if  heated to 350 k  at  1 atm  is  291.67  Ml



<u>calculation</u>

This  is  solved  using  the  Charles  law formula  since the  pressure  is constant.

that is V1/T1 = V2/T2  where,

V1 =250 ml

T1=300  K

V2=?

T2= 350 k


by making V2 the subject   of  the formula  by  multiplying  both side by T2

V2= T2V1/T1

V2= (350 K x 250 ml) / 300K =291.67 Ml


5 0
3 years ago
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