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3 years ago

The discovery of protons and electrons most directly refuted or replace the idea that

Chemistry

1 answer:

Effectus [21]3 years ago

8 0

Answer:

The atom resembles plum pudding

Explanation:

The discovery of the electron in 1897 by J. J Thomson and the proton in 1917 by Rutherford most directly refuted or replace the idea that the atom resembles plum pudding.

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: Citric acid, H3C6H5O7, is a triprotic acid. Consider a buffer system comprising H2C6H5O7 - and HC6H5O7 2- ions. What is the ne

olchik [2.2K]

Answer:

The Net reaction is

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
$HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH$

Explanation:

From the Question we are told that the buffers are

$H_2C_6H_5O_7^ -$ and $HC_6H_5O_7 ^{ 2-}$

When NaOH is added the Net ionic reaction would be

$H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction$
$H_2C_6H_5O_7^ - + OH^- ----> H_C_6H_5O_7^{ 2-} + H_2O \ \ \ \ Add \ NaOH$
$HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH$

8 0

3 years ago

Read 2 more answers

The following mechanism has been proposed for the conversion of tert-butyl bromide to tert-butyl alcohol in aqueous solution: st

Gelneren [198K]

Answer:

See explanation

Explanation:

The first step in this reaction is a unimolecular reaction. It involves the formation of the carbocation. This is so because tertiary alkyl halides only undergo substitution by SN1 mechanism due to sterric crowding.

The second step in the reaction is bi molecular. In this step, the carbocation now combines with the OH^- to yield the alcohol.

Net equation of the reaction is;

(CH3)3CBr + OH^- -------> (CH3)3COH + Br^-

The intermediate here is the carbocation, (CH3)3C^+

8 0

3 years ago

SOMEONE PLEASE HELPPP

xxTIMURxx [149]

A is the correct awnser Beacuse it ether right kne

5 0

3 years ago

What is the pun when I go from here to there quickly on foot

zlopas [31]

1. A type of mens cologne
2. A police officers boss
3. When i go from here to there quickly on foot

Idk thats what google said

6 0

3 years ago

10g of hyrogen react with excess of oxygen gas according to the equation:

Degger [83]

Answer:

Volume of O₂ = 56 dm³

mass of water vapors (H₂O) = 90 g

Explanation:

Data Given:

mass of Oxygen = 10 g

Volume of Oxygen = ?

mass of the water vapor = ?

Reaction Given:

2H₂+O₂---->2H₂O

Solution:

First we have to look at the reaction for the information required

2H₂ + O₂ -------> 2H₂O

2 mol 1mol 2 mol

now convert moles to grams

molar mass of H₂ = 2(1) = 2 g/mol

molar mass of O₂ = 2(16) = 32 g/mol

molar mass of H₂0 = 2(1) + 16 = 18 g/mol

So the masses will be

2H₂ + O₂ -------> 2H₂O

2 mol (2 g/mol) 1mol (32 g/mol) 2 mol (18 g/mol)

4 g 32 g 36 g

So now we know that

4 g of hydrogen combine with 32 g of Oxygen and give 36 g of water vapors.

By using above information

First we find the volume of Oxygen:

For this first we find mass and then moles of Oxygen

As we know

if 4 g of hydrogen combine with 32 g of Oxygen then how much oxygen will react with 10 g of hydrogen

Apply unity formula

4 g of hydrogen H₂ ≅ 32 g of Oxygen O₂

10 g of hydrogen H₂ ≅ X g of Oxygen O₂

by doing Cross multiplication

g of Oxygen O₂ = 32 g x 10 g / 4 g

g of Oxygen O₂ = 80 g

So,

mass of oxygen = 80 g

now find moles of oxygen

formula used:

no. of moles = mass in grams/ molar mass . . . . . . (1)

Put values in above equation 1

no. of moles = 80 g / 32 g/mol

no. of moles = 2.5

Now to find volume of oxygen

Formula used

Volume of O₂ = no. of moles x molar volume (22.4 dm³/ mol) . . . . . . (2)

Put values in equation 2

Volume of O₂ = 2.5 moles x 22.4 dm³/mol

Volume of O₂ = 56 dm³

______________________

Now to find mass of water vapors

As we now

if 4 g of hydrogen produce 36 g of water vapors then how much water vapor will produce from 10 g of hydrogen

Apply unity formula

4 g of hydrogen H₂ ≅ 36 g of water vapors (H₂O)

10 g of hydrogen H₂ ≅ X g of water vapors (H₂O)

by doing Cross multiplication

g of water vapors (H₂O) = 36 g x 10 g / 4 g

g of water vapors (H₂O) = 90 g

So,

mass of water vapors (H₂O) = 90 g

5 0

3 years ago