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3 years ago

How much energy is needed to completely boil a 5.05g sample of water?

1 answer:

8 0

Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

<h3>Why does it take this much energy to boil the water?</h3>

We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.

Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

To learn more about the behavior of water visit:

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In the Ohm's law formula, the letter E stands for B. voltage.

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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percentage, the student measures out 5.84 grams o

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Answer:

The volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml

Explanation:

Here we have the reaction of AgNO₃ and NaCl as follows;

AgNO₃(aq) + NaCl(aq)→ AgCl(s) + NaNO₃(aq)

Therefore, one mole of silver nitrate, AgNO₃, reacts with one mole of sodium chloride, NaCl, to produce one mole of silver choride, AgCl, and one mole of sodium nitrate, NaNO₃,

Therefore, since 5.84 grams of NaCl which is 58.44 g/mol, contains

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0.09993 moles of NaCl will react with 0.09993 moles of AgNO₃

Also, as 1.0 M solution of AgNO₃ contains 1 mole per 1 liter or 1000 mL, therefore, the volume of AgNO₃ that will contain 0.09993 moles is given as follows;

0.09993 × 1 Liter/mole= 0.09993 L = 99.93 mL

Therefore, the volume of 1.0 AgNO₃ that would be required to precipitate 5.84 grams of NaCl is 99.93 ml.

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4 years ago

At a certain temperature, the percent dissociation (ionization) of chlorous acid, HClO2, in a 1.43 M solution water is 8.0%. Cal

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Answer:

Kₐ = 9.9 × 10⁻³

Explanation:

HClO₂ + H₂O ⇌ H₃O⁺ + ClO₂⁻

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[H₃O⁺] = [ClO₂⁻] = 0.1144

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Kₐ = 9.9 × 10⁻³

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3 years ago

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Answer:

The final solution has a concentration of 5.95 * 10^-6 M

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⇒ If 10 mL is diluted to 100 ml then this is a 10x dilution (100 ml /10 ml = 10) so the final solution will be 10 x more dilute.

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4 years ago

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Answer:

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