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3 years ago

In a chemical formula such as C6H12O6 and H2O, what do you call the little numbers below the normal line of text.

Chemistry

2 answers:

Morgarella [4.7K]3 years ago

3 0

These are called subscript number.

That is the number below the normal line of test are called subscript number.

This number indicate the indicate the number of atoms of the element present in the chemical formula.

In both of these C₆H₁₂O₆ and H₂O, the number written below the line of the text are called subscript numbers.

Firdavs [7]3 years ago

3 0

Answer:

subscript number

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3 years ago

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2 years ago

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25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

weeeeeb [17]

Answer:

pH = 11.95≈12

Explanation:

Remember the reaction among aqueous acetic acid ( $CH_3COOH$ ) and aqueous sodium hydroxide (NaOH)

$CH_3COOH + NaOH -> CH_3COONa + H_2O$

First step. Need to know how much moles of the substances are present

$0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH$ = 0.0025 mol NaOH

0.003 mol NaOH * $1 mol CH_3COOH$ / 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> $Na^{+} + OH^{-}$

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

$pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95$

6 0

3 years ago

The Arrhenius theory explains how the concentration of H

koban [17]

Answer:true

Explanation:

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1 year ago

Please leave workings as well

yulyashka [42]

Answer:

15g

Explanation:

The equation for the reaction is given below:

CaO + SO2 —> CaSO3

Molar Mass of CaO = 40 + 16 = 56g/mol

Molar Mass of CaSO3 = 40 + 32 + (16x3) = 40 + 32 + 48 = 120g/mol

From the equation,

56g of CaO produced 120g of CaSO3.

Therefore, 7g of CaO will produce = (7 x 120)/56 = 15g of CaSO3

Therefore, 15g of CaSO3 is produce

5 0

3 years ago