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Zarrin [17]
3 years ago
7

A gas at 1.5 atm had pits ressure decreased to 0.50 atm producing a new volume of 750 ml what was its original volume

Chemistry
1 answer:
Vlada [557]3 years ago
6 0

Answer:

The original volume of the gas is 0.001 mL

Explanation:

This easy excersise can be solved by the law for gases, about pressure and volume; the volume of a gas is inversely proportional to the pressure it exerts.

We can propose the rule by this formula:

P₁  / V₁ = P₂ / V₂

We replace data given: 1.50 atm / V₁ = 0.50 atm / 750 mL

As the rule says, that volume is inversely proportional, and the pressure was decreased, volume must be lower than 750 mL.

1.5atm / (0.5 atm / 750mL) = V₁

V₁ = 0.001 mL

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Using the equations 2 Sr(s) + O₂ (g) → 2 SrO (s) ∆H° = -1184 kJ/mol SrO (s) + CO₂ (g) → SrCO₃ (s) ∆H° = -234 kJ/mol CO₂ (g) → C(
kkurt [141]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is 72 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:

2SrCO_3(s)\rightarrow 2Sr(s)+2C(s)+3O_2(g)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) 2Sr(s)+O_2(g)\rightarrow 2SrO(s)    \Delta H_1=-1184kJ

(2) SrO(s)+CO_2(g)\rightarrow SrCO_3(s)     \Delta H_2=-234kJ      ( × 2)

(3) CO_2(g)\rightarrow C(s)+O_2(g)     \Delta H_3=394kJ    ( × 2)

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times (\Delta H_1)]+[2\times (-\Delta H_2)]+[2\times (\Delta H_3)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-1184))+(2\times -(-234))+(2\times (394))]=72kJ

Hence, the \Delta H^o_{rxn} for the reaction is 72 kJ.

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3 years ago
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5 0
3 years ago
What is the net ionic equation for the reaction that occurs when aqueous solutions of koh and zncl2 are mixed?
Mademuasel [1]

Answer:

Net ionic equation:

Zn²⁺(aq)  +   2OH⁻(aq)     →    Zn(OH)₂(s)

Explanation:

Chemical equation:

ZnCl₂ + KOH    →   KCl  + Zn(OH)₂

Balanced chemical equation:

ZnCl₂ + 2KOH    →   2KCl  +Zn(OH)₂

Ionic equation;

Zn²⁺(aq)  + 2Cl⁻(aq)  + 2K⁺(aq)  +  2OH⁻(aq)     →   2K⁺(aq)  + 2Cl⁻(aq)   +Zn(OH)₂(s)

Net ionic equation:

Zn²⁺(aq)  +   2OH⁻(aq)     →    Zn(OH)₂(s)

The K⁺  and Cl⁻  are spectator ions that's why these are not written in net ionic equation. The  Zn(OH)₂ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

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3 years ago
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Rashid [163]

Answer: 317 joules

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Thus, Q = MCΦ

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Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

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