All categories

4 years ago

Which characteristic of a strong acid

Chemistry

1 answer:

Volgvan4 years ago

5 0

Answer:

It ionizes completely in solution

Explanation:

The main characteristics of a strong acid is that it ionizes completely in solution. A strong acid will produce hydroxonium ions in solution.

A weak acid does not ionize completely in solution. Their ionization is very small and most times, the exists in equilibrium phase with the product formed.

A weak base also ionizes partially in solution.

A weak acid does not ionize completely in a given solution and it exists in equilibrium with the product.

The actions of strong acids are usually fast.

You might be interested in

10. the answer is not D , I know for a fact

Cerrena [4.2K]

A I have a periodic table with me I checked

8 0

3 years ago

Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)

Simora [160]

Answer: The $K_{eq}$ of the reaction is $1.73\times 10^{16}$

Explanation:

For the given half reactions:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V$

Net reaction: $Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)$

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:

$E^o_{cell}=-0.28-(-0.76)=0.48V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}$

Hence, the $K_{eq}$ of the reaction is $1.73\times 10^{16}$

8 0

3 years ago

What is the empirical formula of an oxide of chromium that is 48% oxygen

AnnyKZ [126]

Answer:

$CrO_3$

Explanation:

Hello,

In this case, considering that the compound is 48% oxygen, we infer it is 52% chromium, therefore, by using their atomic masses we compute the moles by assuming those percentages as masses:

$n_{Cr}=52g*\frac{1mol}{52g}=1mol\\ \\n_O=48g*\frac{1mol}{16g}=3mol$

Next, by dividing by the smallest moles, we find the subscripts in the empirical formula:

$Cr=\frac{1}{1} =1\\\\O=\frac{3}{1} =3$

Which are also expressed in the smallest whole numbers, therefore the empirical formula is:

$CrO_3$

Which corresponds to the chromic oxide or chromium (IV) oxide.

Regards.

6 0

3 years ago

16) How many photons are contained in a burst of yellow light (589 nm) from a sodium lamp that contains 609 kJ of energy?

Len [333]

The correct answer to this question is this one:

find the energy of one photon:

E=h*c/λ

divide the energy given by the energy of one photon of that wavelength

What I've done so far is convert wave length to m and energy to j.

E photon = h * x / wave length
E = (6.626 x 10^-43)(3.00 x 10^8) / 587 ^ -9 = 3.38 x 10 ^18 J
3.38 x 10 ^18 J x 1000 kj / 1 j = 3.37 x 10 ^ 16 Kj
609 kJ/ 3.37 x 10 ^ 16 Kj = 1.81 x 10 ^ 16

E = (6.626 x 10^-34)(3.00 x 10^8) / 587 ^ -9 = 3.38 x 10 ^19 J
3.38 x 10 ^19 J x 1000 kj / 1 j = 3.37 x 10 ^ -16 Kj
609 kJ/ 3.37 x 10 ^ 16 Kj = 1.81 x 10 ^ 18 but the answer is 1.81 × 10^24 photons

3.38 x 10 ^-19 J
should be negative

then 3.38 x 10 ^18 J x 1kJ/1000 J

you're converting from J to kJ.. just like meters to kilometres, you wouldn't multiply you would divide

4 0

3 years ago

Read 2 more answers

Why is Azulene aromatic ?

evablogger [386]

Add 2 that’s why and I said if

3 0

3 years ago