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Zanzabum
3 years ago
11

What is the answers to these questions?

Chemistry
1 answer:
serious [3.7K]3 years ago
4 0

2. B

3. B

4. A

5. C

6. B

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What is the symbol of the ion with 36 electrons and a +2 charge?
AlladinOne [14]

Answer:

Strontium

Explanation:

The atomic number of strontium is 38.

It has 38 electrons.

It is alkaline earth metal. It has two valance electrons.

Strontium loses its two electrons and form cation with +2 charge.

Electronic configuration;

Sr₃₈ = [Kr] 5s²

The valance electrons present in 5s are lost by strontium atom and form Sr⁺² cation.

it is yellowish-white metal.

It is highly reactive.

It form salt with halogens.e.g

Sr    +   Br₂    →     SrBr₂

IT react with oxygen and form oxide.

2Sr   +   O₂   →    2SrO

this oxide form hydroxide when react with water,

SrO  + H₂O   →  Sr(OH)₂

With nitrogen it produced nitride,

3Sr + N₂     →  Sr₃N₂

With acid like HCl,

Sr + 2HCl  →  SrCl₂ + H₂

3 0
3 years ago
What Period # is this element found in? P​
77julia77 [94]
15 is the group that phosphorus is found in.
8 0
3 years ago
Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.
zubka84 [21]

<u>Answer:</u> The value of \Delta G^o of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)

  • The equation used to calculate enthalpy change is of a reaction is:

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

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]

We are given:

\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol

  • The equation used to calculate entropy change is of a reaction is:

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

The equation for the entropy change of the above reaction is:

\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]

We are given:

\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol

Putting values in above equation, we get:

\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}

where,

\Delta H^o_{rxn} = standard enthalpy change of the reaction =-67200 J/mol

\Delta S^o_{rxn} = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol

Hence, the value of \Delta G^o of the reaction is 28.38 kJ/mol

7 0
3 years ago
There are 9.88x1023 molecules of O2 available.
blondinia [14]

Answer:

1.64 moles O₂

Explanation:

Part A:

Remember 1 mole of particles = 6.02 x 10²³ particles

So, the question becomes, how many  '6.02 x 10²³'s are there in 9.88 x 10²³ molecules of O₂?

This implies a division of given number of particles by 6.02 x 10²³ particles/mole.

∴moles O₂ = 9.88 x 10²³ molecules O₂ / 6.02 x 10²³ molecules O₂ · mole⁻¹ = 1.64 mole O₂

_______________

Part B needs an equation (usually a combustion of a hydrocarbon).

7 0
2 years ago
Suppose you put a whole antacid tablet in one glass of water and a crushed antacid tablet in another glass containing the same a
iogann1982 [59]
The crushed tablets would stop bubbling/fuzzing first because it has a smaller surface area which means that it would dissolve before the uncrushed tablets which has a larger surface area.
3 0
3 years ago
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