Ionization of sodium dichromate

Sulfur hexafluoride, SF6, is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 2.02 moles o

miss Akunina [59]

Answer:

9.15 atm

Explanation:

Ideal gas equation of state PV=nRT

P in hPa, V in L, n in mol, R is a constant which is 83.1 hpa*L/mol*k, T in kelvin.

Plug in all the number, and we will get:

P*6.21=2.02*83.1*343

P =9271.6(in hpa)=9.15 atm

8 0

2 years ago

How is glaas made ? I reallyt don't know

Vlad1618 [11]

Glass is formed when sand or rocks are heated to high temperatures and then cooled rapidly to create glass. i learned that from minecraf,t when u put sand in a furnace lol

4 0

3 years ago

Benzene is an organic chemical compound with the molecular formula C6H6. In the benzene molecule, carbon atoms form a ring with

Sergeu [11.5K]

Answer:

$sp^2$

Explanation:

Hello,

The six-carbon benzene ring contains two types of bonds: C-C and C-H bonds, that are $sp^2$ -hybridized σ bonds, and the six π bonds that form the aromatic ring. The σ bonds form from one $s$ orbital and two $p$ orbitals from each carbon, which then bond the carbon to the two carbons on either side and the carbon's single hydrogen. The remaining $p$ orbital from each carbon atom sticks out above and below the plane of the ring; these $p$ orbitals overlap sideways, rather than lengthwise, to form the aromatic π bond system.

Best regards.

8 0

2 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

NeX [460]

Explanation:

Reaction equation is as follows.

$Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)$

Here, 1 mole of $Na_{2}SO_{3}$ produces 2 moles of cations.

$[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58$

= 1.16 M

$[SO^{2-}_{3}] = [Na_{2}SO_{3}]$ = 0.58 M

The sulphite anion will act as a base and react with $H_{2}O$ to form $HSO^{-}_{3}$ and $OH^{-}$ .

As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

= $\frac{10^{-14}}{6.3 \times 10^{-8}}$

= $1.59 \times 10^{-7}$

According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

So, x = $[HSO^{-}_{3}] = [OH^{-}]$ = 0.0003 M

$[SO^{2-}_{3}]$ = 0.58 - 0.0003

= 0.579 M

Now, we will use $[HSO^{-}_{3}]$ = 0.0003 M

The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

$K_{b} = \frac{K_{w}}{K_{a_{1}}}$

= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

Therefore, $7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}$

As, x <<<< 0.0003. So, we can neglect x.

Therefore, $x^{2} = 7.14 \times 10^{-13} \times 0.0003$

= $0.00214 \times 10^{-13}$

x = $0.0146 \times 10^{-6}$

x = $[OH^{-}] = [H_{2}SO_{3}]$ = $1.46 \times 10^{-8}$

$[H^{+}] = \frac{10^{-14}}{[OH^{-}]}$

= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

5 0

3 years ago

A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from

sweet-ann [11.9K]

Answer:

$T_i~=163.1$ ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

Initial temperature of gold= Unknow

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the heat equation:

$Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT$

$Q_A_u=m_A_u*Cp_A_u*deltaT$

We can relate these equations if we take into account that all heat of gold is transfer to the water, so:

$m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT$

Now we can put the values into the equation:

$60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)$

Now we can solve for the initial temperature of gold, so:

$T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20$

$T_i~=163.1$ ºC

I hope it helps!

5 0

3 years ago

Ionization of sodium dichromate​

Ionization of sodium dichromate