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wolverine [178]
3 years ago
15

A scholar measures 20 mL of room temperature water (22ºC) and adds 80 mL of 75°C water to it, following the

Chemistry
1 answer:
Leni [432]3 years ago
4 0

Answer:

Data is not valid

Explanation:

When two liquids having different temperatures are mixed, regardless of the volumes, the final mix temperature will ALWAYS be between the initial temperature values.

1st Law Thermo => Law of Conservation of Energy => Energy can not be created nor destroyed, only changed in form. Mixing 22°C with 75°C will NOT result in a mix having a final temperature of 80°C.

∑ΔE = 0 => (mcΔT)₁ + (mcΔT)₂ = 0

[(20g)(1cal/g·°C)(Tₓ - 22°C)] + [(80g)(1cal/g·°C)(Tₓ - 75°C)] = 0

=> 20(Tₓ - 22) + 80(Tₓ - 75) = 0

=> 20Tₓ - 440 + 80Tₓ - 75 = 0

=> 100Tₓ = 440 + 75 = 515

=> Tₓ = (515/100)°C = 51.5°C final mix temperature

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antiseptic1488 [7]

Answer:

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Explanation:

We are considering that the unknown acid is monoprotic. Let the acid is HA.

The reaction between NaOH and acid will be:

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Thus one mole of acid will react with one mole of base.

The moles of base reacted = molarity of NaOH X volume of NaOH

The volume of NaOH used = Final burette reading - Initial reading

Volume of NaOH used = 22.50-0.55= 21.95 mL

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The molarity of acid will be = \frac{mmole}{volumne(mL)}=\frac{0.33}{25}=0.0132M

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3 years ago
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Hey there!

<span>What activates a convection current, starting the flow of a fluid ?

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<span>D. Temperature differences in the mantle </span><span>
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Explanation:

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