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Vika [28.1K]
3 years ago
15

Old thermometers contained very small amounts of mercury. The mercury in the photo has a melting point of -38.8 degrees Celsius.

What can you conclude about the melting point of the mercury in old thermometers? *
It's melting point can only be determined when the mercury is burned.
It's melting point changes as the temperature of mercury changes.
It's melting point equals -38.8 degrees Celsius because it is mercury.
It's melting point is less than -38.8 degrees Celsius because its volume is smaller.
Chemistry
1 answer:
Nikolay [14]3 years ago
7 0

Answer:

••It's melting point equals -38.8 degrees Celsius because it is mercury. |••

Explanation:

This one makes the most sense out of the answers.

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A sample of 250 g of water are heated from 40°C to 121°C, calculate the amount of heat energy absorbed.
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Answer:

the anwser isn't in the choices

Explanation:

H=MC(change of temp.)

M=mass of water=250g

C=specific heat of water = 4.186 j/g

change in temperature is 121-40= 81

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3 years ago
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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