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3 years ago

How much eat is absorbed when 50.0g of water changes from 25.0 Celsius to 37.0 Celsius?

2 answers:

6 0

4.186 j/g. X 50g x (37-25) =
2,511 joules or 2.5 kilo joules
Or 2.38 btu

4.186 is specific heat of water per gram per deg C at sea level
Times mass of 50 grams
Times delta T of the water

2530 joules when cross checked by plugging into an online calculator

Firdavs [7]3 years ago

5 0

27 I think beacuse if you subtract

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What do acids do in solution?

KiRa [710]

B
The answer should be B.....

3 0

2 years ago

Name 2 separate technique in separating iodine crystal and iron filings

KonstantinChe [14]

One is through sublimation, where the mixture is heated and iodine gets converted into gaseous form, leaving behind the iron fillings.

The other is to get a magnet near the mixture and all the iron fillings get attracted to it while iodine will be left over

5 0

3 years ago

Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre

weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

$\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)$

which is equivalent to

$\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)$

The question states that the second equation has an enthalpy, or "heat", of neutralization of $-56.2 \; \text{kJ}$ . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce $56.2 \; \text{kJ}$ or $56.2 \times 10^{3}\; \text{J}$ of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release $0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J}$ of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or $1.00 \times 10^{3} \; \text{ml}$ of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of $1.00 \times 10^{3} \; \text{g}$ .

The solution has a specific heat of $4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . The solution thus have a heat capacity of $4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}$ . Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of $4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}$ , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. $1.405 \times 10^{4} \; \text{J}$ are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0

3 years ago

If the trend in a property is periodic, it means it will

kotegsom [21]

Answer:

d) repeat

Explanation:

If the trend in a property is periodic, it means it will repeat on the periodic table.

Periodic properties on the table have a constant pattern as we move up or down a group or across a period from left to right.

This helps to predict some of the salient properties of elements as we move through the periodic table.
For example, on most periodic groups, metallicity increases as we move down the group and it decreases across the period.

8 0

3 years ago

5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the

cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

$q=mc\Delta T$