Os

Planck’s constant relates the joules of energy absorbed/released by matter to the ________. atomic number of the element frequen

mamaluj [8]

Planck’s constant relates the joules of energy absorbed/released by matter to the wave frequency f. the plancks constant was first recognized in 1900 by Max Planck. The equation that relates the joules of energy absorbed/released by matter to the wave frequency f is called the plancks-eintein relation, E = hf

5 0

3 years ago

Give the expression for the solubility product constant for baf2.

svetoff [14.1K]

Solubility product constant (Ksp) is applied to the saturated ionic solutions which are in equilibrium with its solid form. The solid is partially dissociated into its ions.

For the BaF, the dissociation as follows;
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Hence,
Ksp = [Ba²⁺(aq)] [F⁻(aq)]²

3 0

3 years ago

2C_H. + 702 — 400, + 6H2O

astra-53 [7]

Balanced Eqn

2

C

2

H

6

+

7

O

2

=

4

C

O

2

+

6

H

2

O

By the Balanced eqn

60g ethane requires 7x32= 224g oxygen

here ethane is in excess.oxygen will be fully consumed

hence

300g oxygen will consume

60

⋅

300

224

=

80.36

g

ethane

leaving (270-80.36)= 189.64 g ethane.

By the Balanced eqn

60g ethane produces 4x44 g CO2

hence amount of CO2 produced =

4

⋅

44

⋅

80.36

60

=

235.72

g

and its no. of moles will be

235.72

44

=5.36 where 44 is the molar mass of Carbon dioxide

hope this helps

6 0

3 years ago

You have a 250. -ml sample of 1. 28 m acetic acid (ka = 1. 8 x´ 10–5). calculate the ph of the best buffer.

Nadya [2.5K]

The ph of the best buffer is 4.74

The given acetic acid is a weak acid

The equation of the pH of the buffer

pH = pKa + log ( conjugate base / weak acid ).

For best buffer the concentration of the weak acid and its conjugate base is equal.

pH = pKa + log 1

pH = pKa + 0

pH = pKa

given Ka = 1.8 × 10⁻⁵

pKa = - log ka

pH = -log ( 1.8 × 10⁻⁵ )

pH = 4. 74

Hence the pH of the best buffer is 4.74

Learn more about the pH on

brainly.com/question/22390063

#SPJ4

4 0

2 years ago

Read 2 more answers

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$