Which statement is true about oxygen, sulfur, and selenium pls help :’)

Please answer truthfully:))

makvit [3.9K]

Answer:

Fast, Direction, Time

Explanation:

4 0

2 years ago

What did J.J. Thomson discover about the atom that changed the atomic model previously used?

galina1969 [7]

J.J. Thomson hypothesized and discovered that the atom was not the smallest unit of matter but that instead there were much smaller units. He discovered "sub-atomic particles" which make up atoms. The sub-atomic particle that Thomson discovered was the electron. He discovered this through a process of experiments testing cathode rays.

7 0

3 years ago

Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul

BlackZzzverrR [31]

Answer:

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Explanation:

Assuming that all caculations are at normal pressure and -1.72°C :

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Where

$n$ is the number of moles of hydrogen

$n$ is the mass of hydrogen

$\rho$ is the density of hydrogen

6 0

3 years ago

Read 2 more answers

How does an indicator differentiate between an acid and base?

Svet_ta [14]

Answer:

that is all I know sorry hope that may help you :)

Explanation:

Indicator is a substance which shows different colors in acidic and basic medium. Indicators can be natural (derived from natural sources) or artificial ( man-made) for example :

*Litmus is an indicator. Acid turns blue litmus into red while base turns red litmus into blue

* Turmeric solution does not show change in color (remains yellow) in acidic solution and turns red in basic solution.

8 0

2 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

<u>Answer:</u> The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

<u>Explanation:</u>

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

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

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago