Most polyatomic ions are cations.<br> TRUE<br> FALSE

Information about tectonic plate boundaries has been primarily found by scientists studying data from

taurus [48]

Studying data from earthquakes!

8 0

3 years ago

Read 2 more answers

Get it right :D if this is wrong or not correct ill give u a bad review And no brainiest

tangare [24]

Answer:

In short, because we are tearing up the oxygen factories to make way for carbon dioxide emitters. (Doesn't make a lot of sense, read the explanation)

Explanation:

So, 1000 years ago, we had a lot more trees, didn't have engines or cars or factories or anything, really that released carbon dioxide into the air and we had a lot more trees and since the invention of cars, engines, carbon dioxide-emitting tools and factories and all the other things that emit "Greenhouse gases" and in doing that, cleared more trees to make room for factories and roads and that has drastically changed the outlook of the carbon cycle.

4 0

3 years ago

Drag and drop the labels provided into the boxes below to explain how the Earth-Sun-Moon system results in these seasons on Eart

GalinKa [24]

In order: tilted, rotates, revolving.

6 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

6. What is the oxidation number for the atom indicated in the following compounds.

Hatshy [7]

Answer:

a. +6;

b. +5;

c. +3.

Explanation:

Start with elements with well-known oxidation states.

The oxidation state on oxygen O in compounds is mostly -2. Common exceptions include:

-1 in peroxides and
positive when oxygen bonds to fluorine.

The oxidation state on group 1 metals (Li, Na, K, etc.) in compounds is mostly +1.

The oxidation state on group 2 metals (Be, Mg, Ca, etc.) in compounds is mostly +2.

Barium Ba is a group 2 metal. The oxidation state on Ba in the compound BaSO₄ is expected to be +2.

The oxidation state on hydrogen H in compounds is mostly +1. The oxidation state on H might be negative when it is bonded to metals.

The oxidation state on halogens (F, Cl, Br, etc.) is mostly -1. The oxidation state may vary when the halogen is bonded to oxygen or another halogen element.

Compounds are neutral. The oxidation state on all atoms in a compound shall add up to 0. Both BaSO₄ and HClO₂ are neutral.

Oxidation states:

Ba: +2;
The oxidation state on sulfur S is to be determined;
O: -2.

Let the oxidation state on S be x.

2 + x + 4 × (-2) = 0;

x = 6.

Hence, the oxidation state on S in BaSO₄ is +6.

Oxidation states:

H: +1;
Cl here is bonded to oxygen. The oxidation state on chlorine Cl is to be determined;
O: -2.

Let the oxidation state on Cl be x.

<em>Refer to the equation in BaSO₄ as an example. Try setting up the equation on your own. </em>

x = 3.

Hence, the oxidation state on Cl is +3.

Ions carry charge. Oxidation states on atoms in an ion shall add up to the charge of the ion. The superscript of an ion shows its charge. The superscript 3- in the phosphate ion shows that the ion carries a charge of -3.

Oxidation states:

The oxidation state on P is to be found;
O: -2.

Let the oxidation state on P be x.

x + 4 × (-2) = -3;

x = 5.

Hence, the oxidation state on P is +5.

4 0

3 years ago

Most polyatomic ions are cations. TRUE FALSE