What is a stock solution?

List first thirty elements with their valences

lbvjy [14]

Element Atomic Number Valency

Valency of Hydrogen 1 1

Valency of Helium 2 0

Valency of Lithium 3 1

Valency of Beryllium 4 2

Valency of Boron 5 3

Valency of Carbon 6 4

Valency of Nitrogen 7 3

Valency of Oxygen 8 2

Valency of Fluorine 9 1

Valency of Neon 10 0

Valency of Sodium (Na) 11 1

Valency of Magnesium (Mg) 12 2

Valency of Aluminium 13 3

Valency of Silicon 14 4

Valency of Phosphorus 15 3

Valency of Sulphur 16 2

Valency of Chlorine 17 1

Valency of Argon 18 0

Valency of Potassium (K) 19 1

Valency of Calcium 20 2

Valency of Scandium 21 3

Valency of Titanium 22 4

Valency of Vanadium 23 5,4

Valency of Chromium 24 2

Valency of Manganese 25 7, 4, 2

Valency of Iron (Fe) 26 2, 3

Valency of Cobalt 27 3, 2

Valency of Nickel 28 2

Valency of Copper (Cu) 29 2, 1

Valency of Zinc 30 2

5 0

2 years ago

Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag. 2N

Vitek1552 [10]

Answer : The value of work done for the system is 1144.69 J

Explanation :

First we have to calculate the moles of $NaN_3$

$\text{Moles of }NaN_3=\frac{\text{Mass of }NaN_3}{\text{Molar mass of }NaN_3}$

Molar mass of $NaN_3$ = 65.01 g/mole

$\text{Moles of }NaN_3=\frac{20.2g}{65.01g/mole}=0.311mole$

Now we have to calculate the moles of nitrogen gas.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 2 mole of $NaN_3$ react to give 3 mole of $N_2$

So, 0.311 moles of $NaN_3$ react to give $\frac{0.311}{2}\times 3=0.466$ moles of $N_2$

Now we have to calculate the volume of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = ?

n = number of moles $N_2$ = 0.466 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $22^oC=273+22=295K$

Putting values in above equation, we get:

$1.00atm\times V=0.466mole\times (0.0821L.atm/mol.K)\times 295K$

$V=11.3L$

As initially no nitrogen was present. So,

Volume expanded = Volume of nitrogen evolved

Thus,

Expansion work = Pressure × Volume

Expansion work = 1.00 atm × 11.3 L

Expansion work = 11.3 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 11.3 × 101.3 = 1144.69 J

Therefore, the value of work done for the system is 1144.69 J

5 0

2 years ago

Is iron bromide magnetic if no why

Nataly_w [17]

Iron bromide isn't considered magnetic because all iron compounds are not magnetic

4 0

2 years ago

Read 2 more answers

Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

4 0

2 years ago

Chile~ i need help once again

tensa zangetsu [6.8K]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers