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Anit [1.1K]
4 years ago
12

Determine the oxidation number of the element that is underlined within the compound.

Chemistry
1 answer:
sergey [27]4 years ago
5 0
<h2>Determine the oxidation number of the element that is underlined within the compound. </h2><h2>Li₂ SiO₃ </h2><h2>+2 </h2><h2>+4 </h2><h2>-2 </h2><h2>-4</h2>

Explanation:

Oxidation number

It is defined as “residual charge that is present on atom when the atom is in combined state with other atoms”.

Rules to assign and calculate oxidation number  

1. The oxidation number of atoms in their elemental state is taken as zero.  

2. The oxidation number of mono-atomic atoms like Na+ etc is taken as 1.  

3. The oxidation number of Hydrogen is +1 when present with non metals and -1 when present with metals.  

4. The oxidation number of oxygen is -2 in most of the compounds but in peroxides it is -1.

5. The metals always have oxidation number in positive and non metal in negative when present together in ionic compounds.

6. In compounds that have two atoms with different electro negativities, the oxidation number of more electronegative is taken as –ve and for less electronegative it is taken as positive. For example: In OF2 the oxidation number of oxygen will be in positive and oxidation of fluorine will be in negative.  

7. In neutral compounds, the sum of all oxidation numbers is equal to zero.

8. In complex ions, the sum of oxidation states of all the atoms is equal to the charge present on the complex.  

So, following these rules we can find the oxidation states for all the elements above :

The compound given is :

Li₂SiO₃

Let us find for lithium first :

x      +4       -2

Li₂    Si       O₃

2x + (4) + -6 =0

2x-2 =0

2x =2

x=2/2

x=+1

Let us find the oxidation state for Silicon

+1         x     -2

Li₂       Si       O₃

+2 + x -6 =0

x-4 =0

x =+4

Let us find for oxygen

+1       +4       x

Li₂     Si      O₃

2+4 +3x =0

6 + 3 x =0

6=3x

or , x=6/3 =+2

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What is the correct name of this isotope? *<br><br> oxygen - 16<br> oxygen - 8<br> oxygen - 24
artcher [175]

The correct name of this isotope : Oxygen - 16

<h3>Further explanation</h3>

Given

Isotope : ₈¹⁶O

Required

The correct name

Solution

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  

Atomic mass is the average atomic mass of all its isotopes  

In the following element notation,

\large {{{A} \atop {Z}} \right X}

X = symbol of elemental atom

A = mass number

Z = atomic number

The isotope name is usually followed by its mass number, so the symbol above can be expressed as oxygen - 16

4 0
3 years ago
A chemist is using a solution of HNO3 that has a pH of 3.75.
elena-s [515]
What you have to keep in mind is that 
pH   =  - log[H+]
so 3.75 = -log[H+], we get [H+]=10^{-3.75},
that is [H+]≈1.77827941×10^{-4}(i see this result on calculator)
this is just what you want!

5 0
4 years ago
Read 2 more answers
What is the mass of naoh that would have to be added to 500 ml of a solution of 0.20 m acetic acid in order to achieve a ph of 5
il63 [147K]

The correct answer is 2.5 g.

Let x be the mass of NaOH in grams

The molar mass of NaOH = 40.0 gms/mol

Moles of NaOH = mass / molecular mass of NaOH

= x grams / 40.0 gms /mol = 0.025 x mol

Initial moles of CH₃COOH = volume × concentration of CH₃COOH

= 500 / 1000 × 0.20 = 0.1 mol

CH₃COOH + NaOH ⇒ CH₃COONa + H2O

Moles of CH₃COOH left = initial moles of CH₃COOH - moles of NaOH = 0.1 - 0.025x mol

Moles of CH₃COONa formed = moles of NaOH = 0.025x mol

Henderson-Hasselbalch equation:

pH = pKa + log ([CH₃COONa] / [CH₃COOH])

= pKa + log (moles of CH₃COONa /moles of CH₃COOH)

5.0 = 4.76 + log [0.025a / (0.1 - 0.025a)]

log [0.025a / (0.1 - 0.025a)] = 0.24

0.025a / (0.1 - 0.025a) = 10^0.24 = 1.738

0.068445a = 0.17378

a = 0.17378 / 0.068445

= 2.54 g

Mass of NaOH = a = 2.54 g or 2.5 g

4 0
3 years ago
The formula for calgon
Sonbull [250]

Question

The formula for calgon

Answer: Molecular formula of calgon =  Na₆o₁₈P₆

Explanation:

8 0
3 years ago
g A chemist combines 59.9 mL of 0.282 M potassium bromide with 15.4 mL of 0.512 M silver nitrate. (a) How many grams of silver b
Rufina [12.5K]

Answer:

m_{AgBr}=1.48gAgBr

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

KBr(aq)+AgNO_3(aq)\rightarrow AgBr(s)+KNO_3(aq)

Thus, since the potassium bromide and silver nitrate are in a 1:1 mole ratio, the first step is to identify the limiting reactant, by considering the reacting volumes of reactants in order to compute the available moles of potassium bromide and the moles of potassium bromide consumed by the 15.4 mL of 0.512-M solution of silver nitrate:

n_{KBr}=0.0599L*0.282\frac{molKBr}{L} =0.0169molKBr\\\\n_{KBr}^{consumed}=0.0154L*0.512\frac{molAgNO_3}{L} *\frac{1molKBr}{1molAgNO_3}=0.00788molKBr

In such a way, since less moles are consumed than available, we infer that silver nitrate is the limiting reactant, for which the resulting grams of silver bromide (molar mass 187.8 g/mol) result:

m_{AgBr}=0.00788molAgNO_3*\frac{1molAgBr}{1molAgNO_3} *\frac{187.8gAgBr}{1molAgBr} \\\\m_{AgBr}=1.48gAgBr

Best regards.

6 0
4 years ago
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