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kakasveta [241]
2 years ago
8

What is easier to observe about an object,it’s physical properties or it’s chemical properties. Explain why.

Chemistry
1 answer:
Sladkaya [172]2 years ago
7 0

Answer:

Physical Properties

Explanation:

You are able to see physical properties but are unable to see chemical properties.

You might be interested in
What does not add co2 to the atmosphere
marissa [1.9K]

Answer:

Abiotic objects

Ex. bed, table, lamp, chair, blanket, etc.

Abiotic means Non-living.

7 0
2 years ago
Cierto elemento presenta dos isótopos cuyos números de masa suman 110 y cuyos números de neutrones suman 58 ¿Cuál es el número a
Anettt [7]

Answer:

El número atómico de cada uno de los átomos es 26

Explanation:

El número de masa es la suma de las masas del protón y el neutrón de un átomo.

El número atómico es el número de protones en el átomo.

Los parámetros dados son;

La suma del número másico de ambos átomos = 110

La suma de los neutrones = 58

Por lo tanto, sea el número de protones y neutrones en un isótopo = P₁ y N₁ y el número de protones y neutrones en el otro isótopo = P₂ y N₂

Tenemos;

 P₁ + N₁ + P₂ + N₂ = 110

N₁ + N₂ = 58

Por lo tanto;

P₁ + P₂ = 110 - (N₁ + N₂)

P₁ + P₂ = 110 - 58 = 52

Dado que los isótopos son del mismo elemento, sus protones serán iguales, por lo tanto;

P₁ = P₂

P₁ + P₂ = P₁ + P₁ = 2 × P₁

P₁ + P₂ = 52

2 × P₁ = 52

P₁ = 52/2 = 26 = P₂

El número atómico de ambos átomos es el número de protones en el átomo que es 26.

El número atómico del elemento del átomo es 26

6 0
2 years ago
Nitrogen and hydrogen combine to form ammonia in the Haber process. Calculate (in kJ) the standard enthalpy change ∆H° for the r
Kazeer [188]

Answer:

∆H° rxn = - 93 kJ

Explanation:

Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds  broken minus bonds formed (H according to Hess Law.

We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.

              N₂ (g)   +            3H₂ (g)   ⇒                          2NH₃ (g)

1 N≡N = 1(945 kJ/mol)     3 H-H = 3 (432 kJ/mol)       6 N-H = 6 ( 389 kJ/mol)

∆H° rxn = ∑  H bonds broken  - ∑ H bonds formed

∆H° rxn = [ 1(945 kJ)   + 3 (432 kJ) ] - [ 6 (389 k J]

∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ

be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond,  N≡N,  we have to use this one .

8 0
3 years ago
Where does the equilibrium point occur in a reaction system?
Ymorist [56]
Equilibrium occurs when the rate of the forward reaction is the same as the rate of the reverse reaction. This doesn't necessarily mean the concentrations or pressure are the same on both sides of the equation, only the rates are the same
8 0
3 years ago
Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
3 years ago
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